White continues like this to win the molasses ko.Gérard TAILLE wrote:Matti,
takes
connect at
Difficult position with this sutuation in the upper right corner. I guess you spent a lot of time (like I did myself!) to find such position.
In do not understand why, in Chinese rule (PSK), you cannot play "b"?
What about the sequence 1 to 7 in the diagram above, knowing of course that, between each move, you must insert 4 moves in the right corner ?
In addition, even in japonese rules, a move on "b" seems not a bad move (though a little stupid) because black seems able to win the ko "a" in any case.
Martian problem
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Re: Martian problem
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Re: Martian problem
It is a bad move. He cannot win the ko. The game becomes no result.Gérard TAILLE wrote:Matti,
In addition, even in japonese rules, a move on "b" seems not a bad move (though a little stupid) because black seems able to win the ko "a" in any case.
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Re: Martian problem
Yes Matti of course you are right, I forget to take into account a game with no result. Does that mean that, with japenese rules, the game ends always with no result if the players want to avoid to lose the molasses ko? In this case any of the a,b,c moves seems correct and lead to a no result game, doesn't it ?Matti wrote:It is a bad move. He cannot win the ko. The game becomes no result.Gérard TAILLE wrote:Matti,
In addition, even in japonese rules, a move on "b" seems not a bad move (though a little stupid) because black seems able to win the ko "a" in any case.
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Re: Martian problem
I still do not see how white could win this molasses ko. Starting from your last diagram I continue by:
between 
and 
at 
and white must pass and lose the molasses ko
between and at and white must pass and lose the molasses ko
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Re: Martian problem
No. Black can connect atGérard TAILLE wrote:Yes Matti of course you are right, I forget to take into account a game with no result. Does that mean that, with japenese rules, the game ends always with no result if the players want to avoid to lose the molasses ko? In this case any of the a,b,c moves seems correct and lead to a no result game, doesn't it ?Matti wrote:It is a bad move. He cannot win the ko. The game becomes no result.Gérard TAILLE wrote:Matti,
In addition, even in japonese rules, a move on "b" seems not a bad move (though a little stupid) because black seems able to win the ko "a" in any case.
.
Next black swings back the molasses ko. passNow white may swing the molasses ko back and play a move. Black swings it back and passes. Eventually white runs out of external moves and must also pass. Black also passes and the game ends.
If there is and additional ko with the molasses ko the game can go on forever.
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Re: Martian problem
The problem seems to be a misunderstanding of japonese rules.
In the situation above my understanding is that after black white pass, black pass white, in order to avoid losing the game can claim to play in the molasses ko.
Maybe I am wrong but it looks like the following example:
In this situation do we have to conclude (in japonase rule) that is a winning move or is white allowed to retake the ko after two passes?
In the situation above my understanding is that after black
white pass, black pass white, in order to avoid losing the game can claim to play in the molasses ko.Maybe I am wrong but it looks like the following example:
In this situation do we have to conclude (in japonase rule) that
is a winning move or is white allowed to retake the ko after two passes?-
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Re: Martian problem
I had to revise the situation. White has just captured with circled stone.
Black to play. I think, a is best with japanese rules. b is best with AGA rules (SSK) and c is best with Chinese rules (PSK).
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Re: Martian problem
After two consecutive passes, White can require that play resume, but only with Black to play. Then the resumption might go like this.Gérard TAILLE wrote:The problem seems to be a misunderstanding of japonese rules.
In the situation above my understanding is that after black
= passAnd White is embarrassed.
However, White can claim that the Black stones in the molasses ko are dead. The hypothetical play would go like this.
Maybe I am wrong but it looks like the following example:
In this situation do we have to conclude (in japonase rule) that
In this case ifGérard TAILLE wrote:Oops my last diagramm was not correctly built. Here is a new one.
and 
both pass, White can claim that Black is dead. Hypothetical play could go this way.
= passThe Black stones are dead unless Black to play can make a living group on this board, assuming correct play.
Last edited by Bill Spight on Sun Sep 13, 2020 1:45 pm, edited 1 time in total.
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Visualize whirled peas.
Everything with love. Stay safe.
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Re: Martian problem
Hi Bill,
I am not sure to understand the japonese rule !
In the position above you can see 349 black stones against 8 white stones and I suppose 341 white stones were killed.
After black, pass, pass what is the result ?
Can white claim at retaking the ko, killing a lot of black stones and expecting winning easily the resulting position ?
I am not sure to understand the japonese rule !
In the position above you can see 349 black stones against 8 white stones and I suppose 341 white stones were killed.
After black
, pass, pass what is the result ?Can white claim at retaking the ko, killing a lot of black stones and expecting winning easily the resulting position ?
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Re: Martian problem
I begin to see your idea Matti and I suspect you will manage to distinguish AGA and Chinese rules though this new version seems not to work.Matti wrote:I had to revise the situation. White has just captured with circled stone. Black to play. I think, a is best with japanese rules. b is best with AGA rules (SSK) and c is best with Chinese rules (PSK).
In chinese rule I think a play at "b" is still correct with the following sequence (you have always to insert between each move the molasses ko moves) and you see that white will finally be the first to pass.
Concerning japonese rule I do not see how you will solve the problem because almost any move is correct, leading to a "no result" game (but, for the time being, I am not sure of my understanding of japonese rule !)
In addition the martian problem is supposed to start with a position ignoring all the previous moves. As a consequence you cannot make the assomption that a move is unplayable due to a ko.
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Re: Martian problem
In hypothetical play White can takeGérard TAILLE wrote:Hi Bill,
I am not sure to understand the japonese rule !
In the position above you can see 349 black stones against 8 white stones and I suppose 341 white stones were killed.
After black
Can white claim at retaking the ko, killing a lot of black stones and expecting winning easily the resulting position ?
and fill the ko, so the stone is dead. That means that the 
point is a dame, which means that the rest of the Black stones are in seki, which means that Black has no territory. (OC, Black has 342 prisoners and wins the game, anyway. 
)The Adkins Principle:
At some point, doesn't thinking have to go on?
— Winona Adkins
Visualize whirled peas.
Everything with love. Stay safe.
At some point, doesn't thinking have to go on?
— Winona Adkins
Visualize whirled peas.
Everything with love. Stay safe.
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Re: Martian problem
You didn't state this in the problem.Gérard TAILLE wrote:Matti wrote:
In addition the martian problem is supposed to start with a position ignoring all the previous moves. As a consequence you cannot make the assomption that a move is unplayable due to a ko.
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Re: Martian problem
With japanese rules in hypothetical play the circled white stones can also be calied dead.
The circled black stones would also be dead and also the big white group in upper right corner. Dead stones cannot surround territory, so there is territory for neither white nor black. Dead stones may only be removed from territory, so no stones would be removed from the upper right corner.
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Re: Martian problem
The simple answer is "yes". The reasoning is somewhat subtle. White may try two things.Gérard TAILLE wrote:Hi Bill,
I am not sure to understand the japonese rule !
In the position above you can see 349 black stones against 8 white stones and I suppose 341 white stones were killed.
After black
Can white claim at retaking the ko, killing a lot of black stones and expecting winning easily the resulting position ?
First Approach: After White passes, black passes. White tries to claim that the whole black stones are dead. The problem is that, even though White can capture all black stones in the hypothetical play, Black may play at the upper left corner to create a stone White cannot capture. Thus the Black group is not dead by the rule. This is not the right approach.
Second approach: After White passes, black passes. The game temporarily stops but White can request the resumption of the game. Black will play first, but has nothing to do, and White will eventually capture all Black stones. Now, be careful that Black can keep playing as if almost a new game stared. Depending on the captured stone numbers and komi, Black may still have a chance to win. Anyway, this is a better result than the first approach.
Jaeup Kim
Professor in Physics, Ulsan National Institute of Science and Technology, Korea
Author of the Book "Understanding the Rules of Baduk", available at https://home.unist.ac.kr/professor/juki ... ce&wr_id=5
Professor in Physics, Ulsan National Institute of Science and Technology, Korea
Author of the Book "Understanding the Rules of Baduk", available at https://home.unist.ac.kr/professor/juki ... ce&wr_id=5