scoring at the end
Mark-T
players
to mark the dead stones. Why? What is so difficult about
recognizing
them? Even a novice player is capable of this, and does so routinely
(in live play).
--
Mark
He who has the most bamboo joints, wins.
Curt Welch
> I notice that all the go servers, in the scoring phase, need the
> players
> to mark the dead stones. Why? What is so difficult about
> recognizing
> them? Even a novice player is capable of this, and does so routinely
> (in live play).
That's a good question. I never really understood that about go either and
I'd like to see what the people who understand go say about it. My feeling
was that complex games between humans tend to end before it becomes totally
obvious to the computer who is controlling which stones. For example, I
think it's possible to have white stones, surrounding black stones,
sounding white stones surrounding black stones with none of them secured
with a double-eye. But yet the human players choose not to play it out
because they both think it's obvious what would happen if they did.
I think maybe (but I'm not sure) the issue is if the guy who would win,
placed another stone on the board to secure the area, he is loosing a point
to do so. And as such, will choose to end the game, rather than waste a
point to move into the area. Likewise, the other player, also believing
the area is lost, even if he moves into it first, also has no incentive to
play the stone, because it gives his opponent another point. As such, even
if the areas is not obviously secure to the point that would allow the
computer to score it, the game can still be over if both players agree it's
over.
What I don't understand, is if the players don't agree on what's dead and
not dead, what happens? Does the game restart at that point?
--
Curt Welch http://CurtWelch.Com/
cu...@kcwc.com http://NewsReader.Com/
Robert Jasiek
>What I don't understand, is if the players don't agree on what's dead and
>not dead, what happens? Does the game restart at that point?
That depends on the rules.
Robert Jasiek
<MarkTa...@gmail.com> wrote:
>I notice that all the go servers, in the scoring phase, need the
>players to mark the dead stones. Why? What is so difficult about
>recognizing them?
1) Go servers tend to use rulesets with an optional agreement or a
mandatory confirmation phase rather than without, i.e., where the
players would continue to approach and fill liberties.
2) Determination of life and death can become "arbitrarily" complex,
i.e., it can happen that it is beyond the computer's calculation
ability.
The more interesting question would be: Why are programs so stupid to
fail even in seemingly simple game stop positions that most beginners
can decide easily? Answer: because not enough care and thought has
been invested in the programs... But lost semeais can be pretty
difficult to judge.
Joachim Pimiskern
> I notice that all the go servers, in the scoring phase, need the
> players to mark the dead stones. Why? What is so difficult about
> recognizing them? Even a novice player is capable of this, and
> does so routinely (in live play).
Suppose both players agree to end the game having
played 0 moves yet. Now it's up to the go server to
determine who has won. That's as complicated as playing
go itself. Go is AI-complete.
Regards,
Joachim
-
> Mark-T <MarkTa...@gmail.com> wrote:
>> I notice that all the go servers, in the scoring phase,
>> need the players to mark the dead stones. Why?
>> What is so difficult about recognizing them? Even
>> a novice player is capable of this, and does so routinely
>> (in live play).
cu...@kcwc.com (Curt Welch) wrote:
> That's a good question. I never really understood that about
> go either and I'd like to see what the people who understand
> go say about it.
The "scoring phase" (between game-stop and game-end)
still involves game play. It is left up to player agreements to
determine what is alive, dead & seki. They might be in mutual
error however the game rules allow for this. Similarly, in the
scholastic chess scene if a player claims checkmate and the
opponent does not disagree then the game is "over" & "won"
even if the situation was not a checkmate (or stalemate, etc.).
In learning this game who hasn't mistaken seki vs. dead stones?
> I think maybe (but I'm not sure) the issue is if the guy who would
> win, placed another stone on the board to secure the area, he is
> loosing a point to do so. And as such, will choose to end the
> game, rather than waste a point to move into the area. Likewise,
> the other player, also believing the area is lost, even if he moves
> into it first, also has no incentive to play the stone, because it
> gives his opponent another point. As such, even if the areas is
> not obviously secure to the point that would allow the computer
> to score it, the game can still be over if both players agree it's
> over.
Yes.
> What I don't understand, is if the players don't agree on what's
> dead and not dead, what happens? Does the game restart
> at that point?
This is a serious problem which the servers cannot solve.
That's why clients offer a way to maintain the "bozo" list of
opponents not to play again. On IGS games go to arbitration
which is why some fees are collected to pay for arbitrations.
On scoring disputes one may attempt to disconnect and then
reload but if this occurs more than a few times the game is
marked for arbitration or, according to one interpretation, the
player who keeps disconnecting and reloading too many
times suffers an automatic penalty.
In the heydey, bozo players were sometimes excoriated
and pilloried on this newsgroup. As if it mattered ...
- regards
- jb
-------------------------------------------------------
Twin paradox
http://en.wikipedia.org/wiki/Twin_paradox
-------------------------------------------------------
2nd grade computer test in China
http://funstufftosee.com/frogleaptest.html
-------------------------------------------------------
Nick Wedd
<64049981-daa6-4c0b...@r36g2000prf.googlegroups.com>,
Mark-T <MarkTa...@gmail.com> writes
>I notice that all the go servers, in the scoring phase, need the
>players
>to mark the dead stones. Why? What is so difficult about
>recognizing
>them? Even a novice player is capable of this, and does so routinely
>(in live play).
Novice players are not capable of identifying dead stones reliably in
all positions. Nor for that matter are reasonably strong players. And
no-one has been able to write a program that can do this reliably.
Nick
--
Nick Wedd ni...@maproom.co.uk
-
"Nick Wedd" (Brit killfiler) <ni...@maproom.co.uk> wrote:
> Novice players are not capable of identifying dead stones
> reliably in all positions. Nor for that matter are reasonably
> strong players. And no-one has been able to write a
> program that can do this reliably.
Again, this was subject to misinterpretation. Whether
novice, strong, or computer program, a decision process
will render some form of reliable result. To be a correct
result a score should be in agreement with a Tournament
Director, however the TD is not called into action unless
the players request it. In practice, then, the scores are
not checked with the TD so they are just reliable scores.
Disagreements from onlookers and/or other bystanders
are simply not relevant. In fact, there really isn't anything
to discuss here because only players arrive at their score.
And the computer program does behave according
to the scoring rules it has been programmed to follow.
- regards
- jb
-----------------------------------------------------------
Arithmetic-Geometric Mean of a set of numbers ...
-----------------------------------------------------------
10 point 250:dim M(90):S=10:Err=10^(-S):' -- agm.ub
20 I=0:M(0)=1:while M(I):inc I:input M(I):wend:dec I
30 print using(10,S),fnAG(fnAGA(M,I),fnAGR(M,I)):goto 20
40 fnAG(X,Y):while abs(W-Z)>E:W=(X+Y)/2:Z=sqrt(X*Y)
45 X=W:Y=Z:wend:return(W)
50 fnAGA(M,I):W=0:for J=1 to I:W+=M(J):next:return(W/I)
60 fnAGR(M,I):Z=1:for J=1 to I:Z*=M(J):next:return(Z^(1/I))
-----------------------------------------------------------
Robert Jasiek
<JoachimP...@web.de> wrote:
>Suppose both players agree to end the game having
>played 0 moves yet. Now it's up to the go server to
>determine who has won. That's as complicated as playing
>go itself.
How could that be as complicated as playing go? Even under the
Japanese 1989/2003 Rules it is solvable. I would agree though if you
replaced 0 moves by 1 board play before the game stopping passes. 0 is
so much easier because life and death does not have to be determined
at all; only the existence of territory.
Joachim Pimiskern
> On Fri, 13 Mar 2009 08:31:56 +0100, "Joachim Pimiskern"
> <JoachimP...@web.de> wrote:
>>Suppose both players agree to end the game having
>>played 0 moves yet. Now it's up to the go server to
>>determine who has won. That's as complicated as playing
>>go itself.
>
> How could that be as complicated as playing go? Even under the
> Japanese 1989/2003 Rules it is solvable.
Solvable, yes. As far as I know, only 5*5 go has been solved,
that means, it's known who (in this case black) wins.
> I would agree though if you replaced 0 moves by 1 board play
> before the game stopping passes.
If one had a strong scoring algorithm, he could use
it as a base for a go program: suggest each move to it
and calculate the scoring. The move with the highest
scoring is the move to perform on the real board.
Regards,
Joachim
Nick Wedd
<JoachimP...@web.de> writes
There are three different questions, which may be confused here.
One is "given a position and the fact that both players have passed * :
deduce which groups are dead, and calculate the score".
The second is "given a position, find the best move".
The third is "given a position in which neither player has a move better
than pass * : deduce which groups are dead, and calculate the score".
None of these can be achieved (for rulesets in normal use) by any
existing computer program.
* = "(and the numbers of prisoners, and who moved last, if relevant)"
-
> "Joachim Pimiskern" <JoachimP...@web.de> wrote:
>> Solvable, yes. As far as I know, only 5*5 go has been solved,
>> that means, it's known who (in this case black) wins.
Yes. If 5*5 Go has been solved this means it has
been perfectly scored in all instances. What is the big
problem with perfect endgame scoring for 6*6, 7*7... Go ?
"Nick Wedd" <ni...@maproom.co.uk> wrote:
> There are three different questions, which may be confused here.
> [ ... ]
> The third is "given a position in which neither player has a move better
> than pass * : deduce which groups are dead, and calculate the score".
>
> None of these can be achieved (for rulesets in normal use) by any
> existing computer program.
Ever since the BGA frap Nick Wedd, who did badly, has been
ordered by the BGA not reply or interact. Moreoever he continues
to display inveterate ignorance by use of killfiles and/or filters. Why
doesn't Nick Wedd, or anyone else, supply an example of some
endgame position, where neither player has a move better than pass,
which no computer can score correctly?
- regards
- jb
-------------------------------------------------------------------
http://70.87.245.212/story.php?id=6737
Secret Message Found In Lincoln Pocket Watch
-------------------------------------------------------------------
Robert Sneddon
>Solvable, yes. As far as I know, only 5*5 go has been solved,
>that means, it's known who (in this case black) wins.
A have a vague recollection that 7*7 go has been solved for all
possible games but I can't recall who did it and where I saw the claim
made. I think it was done by simple exhaustion of the legal move space
plus symmetry and rotations with lots and lots of computer hours and a
large lump of database storage. I don't know how they dealt with
multiple kos though.
--
To reply, my gmail address is nojay1 Robert Sneddon
Jeff Nowakowski
>
> A have a vague recollection that 7*7 go has been solved for all
> possible games but I can't recall who did it and where I saw the claim
> made.
I highly doubt that. As far as I know, even 5x5 has only been solved
for the opening positions (see the 2003 paper Solving Go On Small
Boards). The 7x7 result is only based on human analysis for best play
from the opening position. If there were further results since then I'd
be surprised I haven't heard about it. A full computer solution for 7x7
would be epic.
-Jeff
Kirk McElhearn
> The more interesting question would be: Why are programs so stupid to
> fail even in seemingly simple game stop positions that most beginners
> can decide easily? Answer: because not enough care and thought has
> been invested in the programs... But lost semeais can be pretty
> difficult to judge.
GnuGo seems to figure out which stones are dead just fine...
Kirk
--
Read my blog, Kirkville
http://www.mcelhearn.com
Robert Jasiek
<JoachimP...@web.de> wrote:
>> How could that be as complicated as playing go? Even under the
>> Japanese 1989/2003 Rules it is solvable.
>Solvable, yes. As far as I know, only 5*5 go has been solved,
>that means, it's known who (in this case black) wins.
[About scoring the empty board.]
Solvable in a small constant time, I mean.
Robert Jasiek
<fr...@nospam.demon.co.uk> wrote:
> A have a vague recollection that 7*7 go has been solved
For a specific ruleset. (I have not proofread the solution though.)
Robert Jasiek
wrote:
>GnuGo seems to figure out which stones are dead just fine...
How does it do against the Hatsuyoron problems?
Mark-T
> There are three different questions, which may be confused here.
> One is "given a position and the fact that both players
> have passed * : deduce which groups are dead, and
> calculate the score".
> ........
> The third is "given a position in which neither player has
> a move better than pass * : deduce which groups are
> dead, and calculate the score".
What's the difference?
--
Mark
wilemien
> On Fri, 13 Mar 2009 17:56:02 +0000, Robert Sneddon
>
> > A have a vague recollection that 7*7 go has been solved
>
> For a specific ruleset. (I have not proofread the solution though.)
Even 7x7 go is as far as i know not solved.
What is done is that there has been an investigation that under
Japanese or Chinese rules the game is even (leads to jigo a draw) if
the komi is 9
But that optimal play does depend the ruleset.
(But there are optimal plays that lead to draws under both rulesets)
see http://senseis.xmp.net/?7x7ArticleByJDavies
There is also an investigation from Tromp, (or more an Sgf about 7x7 go
() but that was also far from complete.
(but it does invalidate some of the statements in Davies,article I
think)
Think that solving 7x7 go is an interesting subject. but is even now
quite an problem.
(not only computational but also how do you do it?)
and what about the different plays for different rulesets?
I would like to help somebdy who wants to solve it
Robert Jasiek
<wile...@googlemail.com> wrote:
>I would like to help somebdy who wants to solve it
Tell this the computer go mailing list; there might be 2 or 3 more
expert programmers on 7x7.
Nick Wedd
<5ba7e104-8f6b-46bd...@r10g2000prf.googlegroups.com>,
Mark-T <MarkTa...@gmail.com> writes
In the first case, both players have passed but it is possible that they
should not have done. There may be something unsettled on the board.
In the second case, there can be nothing unsettled on the board.