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Martian problem

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Re: Martian problem

Posted: Sun Sep 13, 2020 1:10 am
by Matti
Gérard TAILLE wrote:Matti,
Click Here To Show Diagram Code
[go]$$B
$$ -----------------------
$$ | . X O 3 O O X . X O |
$$ | X X O 4 O O X . X O |
$$ | . X X O O O X X O O |
$$ | X X O . O X X O . O |
$$ | O O O . O X O . O O |
$$ | X X O O O X X O O X |
$$ | . X X X O O O X X X |
$$ | X X . X X 1 O X . X |
$$ | X . 6 . X X O O X . |
$$ | X X . X X 2 X O X X |
$$ -----------------------[/go]
:b5: takes :w2: :b7: connect at :w2:

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.
Click Here To Show Diagram Code
[go]$$W
$$ -----------------------
$$ | . X O . O O X 3 X O |
$$ | X X O O O O X 1 X O |
$$ | . X X O O O X X O O |
$$ | X X O . O X X O 4 O |
$$ | O O O . O X O 2 O O |
$$ | X X O O O X X O O X |
$$ | . X X X O O O X X X |
$$ | X X . X X X O X . X |
$$ | X . O . X X O O X . |
$$ | X X 5 X X X X O X X |
$$ -----------------------[/go]
White continues like this to win the molasses ko.

Re: Martian problem

Posted: Sun Sep 13, 2020 1:28 am
by Matti
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.
It is a bad move. He cannot win the ko. The game becomes no result.

Re: Martian problem

Posted: Sun Sep 13, 2020 3:07 am
by Gérard TAILLE
Matti wrote:
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.
It is a bad move. He cannot win the ko. The game becomes no result.
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 ?

Re: Martian problem

Posted: Sun Sep 13, 2020 3:19 am
by Gérard TAILLE
I still do not see how white could win this molasses ko. Starting from your last diagram I continue by:
Click Here To Show Diagram Code
[go]$$W
$$ -----------------------
$$ | . X O . O O X . X O |
$$ | X X O O O O X . X O |
$$ | . X X O O O X X O O |
$$ | X X O 2 O X X O . O |
$$ | O O O 5 O X O . O O |
$$ | X X O O O X X O O X |
$$ | . X X X O O O X X X |
$$ | X X 3 X X X O X . X |
$$ | X 4 O 6 X X O O X . |
$$ | X X 1 X X X X O X X |
$$ -----------------------[/go]
:w7: between :b4: and :b6:
:b8: at :w1:
and white must pass and lose the molasses ko

Re: Martian problem

Posted: Sun Sep 13, 2020 9:46 am
by Matti
Gérard TAILLE wrote:
Matti wrote:
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.
It is a bad move. He cannot win the ko. The game becomes no result.
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 ?
No. Black can connect at :b1: .
Click Here To Show Diagram Code
[go]$$B
$$ -----------------------
$$ | . X O . O O X 4 X O |
$$ | X X O . O O X 2 X O |
$$ | . X X O O O X X O O |
$$ | X X O . O X X O 5 O |
$$ | O O O . O X O 3 O O |
$$ | X X O O O X X O O X |
$$ | . X X X O O O X X X |
$$ | X X . X X 6 O X . X |
$$ | X . . . X X O O X . |
$$ | X X . X X 1 X O X X |
$$ -----------------------[/go]
Next black swings back the molasses ko.
Click Here To Show Diagram Code
[go]$$B
$$ -----------------------
$$ | . X O . O O X O 3 O |
$$ | X X O . O O X O 1 O |
$$ | . X X O O O X X O O |
$$ | X X O . O X X 2 X O |
$$ | O O O . O X 4 X O O |
$$ | X X O O O X X O O X |
$$ | . X X X O O O X X X |
$$ | X X . X X O O X . X |
$$ | X . . . X X O O X . |
$$ | X X . X X X X O X X |
$$ -----------------------[/go]
:b5: pass

Now 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.

Re: Martian problem

Posted: Sun Sep 13, 2020 10:46 am
by Gérard TAILLE
The problem seems to be a misunderstanding of japonese rules.
Click Here To Show Diagram Code
[go]$$B
$$ -----------------------
$$ | . X O . O O X O . O |
$$ | X X O O O O X O . O |
$$ | . X X O O O X X O O |
$$ | X X O O O X X O 1 O |
$$ | O O O . O X . X O O |
$$ | X X O O O X X O O X |
$$ | . X X X O O O X X X |
$$ | X X X X X O O X . X |
$$ | X . X . X X O O X . |
$$ | X X . X X X X O X X |
$$ -----------------------[/go]
In the situation above my understanding is that after black :b1: 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:
Click Here To Show Diagram Code
[go]$$B
$$ ------------
$$ | 1 O O . O|
$$ | O X O O .|
$$ | X X X O O|
$$ | . X X X X|
$$ | X X X X X|
$$ ------------[/go]
In this situation do we have to conclude (in japonase rule) that :b1: is a winning move or is white allowed to retake the ko after two passes?

Re: Martian problem

Posted: Sun Sep 13, 2020 11:05 am
by Matti
I had to revise the situation. White has just captured with circled stone.
Click Here To Show Diagram Code
[go]$$B
$$ -----------------------
$$ | . X O . O O X . X O |
$$ | X X O c . O X . X O |
$$ | . X X O O O X X O O |
$$ | X X O . O X X O . O |
$$ | O O O . O X W . O O |
$$ | X X O O O X X O O X |
$$ | . X X X O O O X X X |
$$ | X X . X X b O X . X |
$$ | X . . . X X O O X . |
$$ | X X . X X a X O X X |
$$ -----------------------[/go]
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).

Re: Martian problem

Posted: Sun Sep 13, 2020 11:09 am
by Gérard TAILLE
Oops my last diagramm was not correctly built. Here is a new one.
Click Here To Show Diagram Code
[go]$$B
$$ ----------------
$$ | 1 O O O . O .|
$$ | O X X O O O O|
$$ | X X X X O O O|
$$ | X X X X O X X|
$$ | X X X O O X .|
$$ | X X X O X . X|
$$ | . X X O X X .|
$$ ------------[/go]

Re: Martian problem

Posted: Sun Sep 13, 2020 12:31 pm
by Bill Spight
Gérard TAILLE wrote:The problem seems to be a misunderstanding of japonese rules.
Click Here To Show Diagram Code
[go]$$B
$$ -----------------------
$$ | . X O . O O X O . O |
$$ | X X O O O O X O . O |
$$ | . X X O O O X X O O |
$$ | X X O O O X X O 1 O |
$$ | O O O . O X . X O O |
$$ | X X O O O X X O O X |
$$ | . X X X O O O X X X |
$$ | X X X X X O O X . X |
$$ | X . X . X X O O X . |
$$ | X X . X X X X O X X |
$$ -----------------------[/go]
In the situation above my understanding is that after black :b1: white pass, black pass white, in order to avoid losing the game can claim to play in the molasses ko.
After two consecutive passes, White can require that play resume, but only with Black to play. Then the resumption might go like this.
Click Here To Show Diagram Code
[go]$$B
$$ -----------------------
$$ | . X O . O O X O 3 O |
$$ | X X O O O O X O 1 O |
$$ | . X X O O O X X O O |
$$ | X X O O O X X 2 X O |
$$ | O O O . O X 4 X O O |
$$ | X X O O O X X O O X |
$$ | . X X X O O O X X X |
$$ | X X X X X O O X . X |
$$ | X . X 5 X X O O X . |
$$ | X X . X X X X O X X |
$$ -----------------------[/go]
:w6: = pass
Click Here To Show Diagram Code
[go]$$B
$$ -----------------------
$$ | . X O . O O X . X O |
$$ | X X O O O O X . X O |
$$ | . X X O O O X X O O |
$$ | X X O O O X X O . O |
$$ | O O O . O X O 7 O O |
$$ | X X O O O X X O O X |
$$ | . X X X O O O X X X |
$$ | X X X X X O O X . X |
$$ | X . X X X X O O X . |
$$ | X X . X X X X O X X |
$$ -----------------------[/go]
And White is embarrassed.

However, White can claim that the Black stones in the molasses ko are dead. The hypothetical play would go like this.
Click Here To Show Diagram Code
[go]$$W Black dies, hypothetically
$$ -----------------------
$$ | . X O . O O X O . O |
$$ | X X O O O O X O 2 O |
$$ | . X X O O O X X O O |
$$ | X X O O O X X 1 X O |
$$ | O O O . O X 3 X O O |
$$ | X X O O O X X O O X |
$$ | . X X X O O O X X X |
$$ | X X X X X O O X . X |
$$ | X . X . X X O O X . |
$$ | X X . X X X X O X X |
$$ -----------------------[/go]
Maybe I am wrong but it looks like the following example:
Click Here To Show Diagram Code
[go]$$B
$$ ------------
$$ | 1 O O . O|
$$ | O X O O .|
$$ | X X X O O|
$$ | . X X X X|
$$ | X X X X X|
$$ ------------[/go]
In this situation do we have to conclude (in japonase rule) that :b1: is a winning move or is white allowed to retake the ko after two passes?
Gérard TAILLE wrote:Oops my last diagramm was not correctly built. Here is a new one.
Click Here To Show Diagram Code
[go]$$B
$$ ----------------
$$ | 1 O O O . O .|
$$ | O X X O O O O|
$$ | X X X X O O O|
$$ | X X X X O X X|
$$ | X X X O O X .|
$$ | X X X O X . X|
$$ | . X X O X X .|
$$ ------------[/go]
In this case if :w2: and :b3: both pass, White can claim that Black is dead. Hypothetical play could go this way.
Click Here To Show Diagram Code
[go]$$W
$$ ----------------
$$ | X O O O . O .|
$$ | 1 X X O O O O|
$$ | X X X X O O O|
$$ | X X X X O X X|
$$ | X X X O O X .|
$$ | X X X O X . X|
$$ | 3 X X O X X .|
$$ ------------[/go]
:b2: = pass
Click Here To Show Diagram Code
[go]$$W
$$ ----------------
$$ | . O O O . O .|
$$ | O . . O O O O|
$$ | . . . . O O O|
$$ | . . . . O X X|
$$ | . . . O O X .|
$$ | . . . O X . X|
$$ | O . . O X X .|
$$ ------------[/go]
The Black stones are dead unless Black to play can make a living group on this board, assuming correct play.

Re: Martian problem

Posted: Sun Sep 13, 2020 1:24 pm
by Gérard TAILLE
Hi Bill,

I am not sure to understand the japonese rule !
Click Here To Show Diagram Code
[go]$$B
$$ ----------------------------------------
$$ | X X X X X X X X X X X X X X X X X X X|
$$ | X X X X X X X X X X X X X X X X X X X|
$$ | X X X X X X X X X X X X X X X X X X X|
$$ | X X X X X X X X X X X X X X X X X X X|
$$ | X X X X X X X X X X X X X X X X X X X|
$$ | X X X X X X X X X X X X X X X X X X X|
$$ | X X X X X X X X X X X X X X X X X X X|
$$ | X X X X X X X X X X X X X X X X X X X|
$$ | X X X X X X X X X X X X X X X X X X X|
$$ | X X X X X X X X X X X X X X X X X X X|
$$ | X X X X X X X X X X X X X X X X X X X|
$$ | X X X X X X X X X X X X X X X X X X X|
$$ | X X X X X X X X X X X X X X X X X X .|
$$ | X X X X X X X X X X X X X X X X X X X|
$$ | X X X X X X X X X X X X X X X X X X O|
$$ | X X X X X X X X X X X X X X X X X O 1|
$$ | X X X X X X X X X X X X X X X X X O O|
$$ | X X X X X X X X X X X X X X X X O O .|
$$ | X X X X X X X X X X X X X X X X O . O|
$$ ----------------------------------------[/go]
In the position above you can see 349 black stones against 8 white stones and I suppose 341 white stones were killed.
After black :b1:, 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 ?

Re: Martian problem

Posted: Sun Sep 13, 2020 1:46 pm
by Gérard TAILLE
Matti wrote:I had to revise the situation. White has just captured with circled stone.
Click Here To Show Diagram Code
[go]$$B
$$ -----------------------
$$ | . X O . O O X . X O |
$$ | X X O c . O X . X O |
$$ | . X X O O O X X O O |
$$ | X X O . O X X O . O |
$$ | O O O . O X W . O O |
$$ | X X O O O X X O O X |
$$ | . X X X O O O X X X |
$$ | X X . X X b O X . X |
$$ | X . . . X X O O X . |
$$ | X X . X X a X O X X |
$$ -----------------------[/go]
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).
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.

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)
Click Here To Show Diagram Code
[go]$$B
$$ -----------------------
$$ | . X O 4 O O X . X O |
$$ | X X O 3 . O X . X O |
$$ | . X X O O O X X O O |
$$ | X X O 7 O X X O . O |
$$ | O O O . O X O . O O |
$$ | X X O O O X X O O X |
$$ | . X X X O O O X X X |
$$ | X X 8 X X 1 O X . X |
$$ | X 9 2 . X X O O X . |
$$ | X X 6 X X 5 X O X X |
$$ -----------------------[/go]
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.

Re: Martian problem

Posted: Sun Sep 13, 2020 3:41 pm
by Bill Spight
Gérard TAILLE wrote:Hi Bill,

I am not sure to understand the japonese rule !
Click Here To Show Diagram Code
[go]$$B
$$ ----------------------------------------
$$ | X X X X X X X X X X X X X X X X X X X|
$$ | X X X X X X X X X X X X X X X X X X X|
$$ | X X X X X X X X X X X X X X X X X X X|
$$ | X X X X X X X X X X X X X X X X X X X|
$$ | X X X X X X X X X X X X X X X X X X X|
$$ | X X X X X X X X X X X X X X X X X X X|
$$ | X X X X X X X X X X X X X X X X X X X|
$$ | X X X X X X X X X X X X X X X X X X X|
$$ | X X X X X X X X X X X X X X X X X X X|
$$ | X X X X X X X X X X X X X X X X X X X|
$$ | X X X X X X X X X X X X X X X X X X X|
$$ | X X X X X X X X X X X X X X X X X X X|
$$ | X X X X X X X X X X X X X X X X X X .|
$$ | X X X X X X X X X X X X X X X X X X X|
$$ | X X X X X X X X X X X X X X X X X X O|
$$ | X X X X X X X X X X X X X X X X X O 1|
$$ | X X X X X X X X X X X X X X X X X O O|
$$ | X X X X X X X X X X X X X X X X O O .|
$$ | X X X X X X X X X X X X X X X X O . O|
$$ ----------------------------------------[/go]
In the position above you can see 349 black stones against 8 white stones and I suppose 341 white stones were killed.
After black :b1:, 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 ?
Click Here To Show Diagram Code
[go]$$B
$$ ----------------------------------------
$$ | X X X X X X X X X X X X X X X X X X X|
$$ | X X X X X X X X X X X X X X X X X X X|
$$ | X X X X X X X X X X X X X X X X X X X|
$$ | X X X X X X X X X X X X X X X X X X X|
$$ | X X X X X X X X X X X X X X X X X X X|
$$ | X X X X X X X X X X X X X X X X X X X|
$$ | X X X X X X X X X X X X X X X X X X X|
$$ | X X X X X X X X X X X X X X X X X X X|
$$ | X X X X X X X X X X X X X X X X X X X|
$$ | X X X X X X X X X X X X X X X X X X X|
$$ | X X X X X X X X X X X X X X X X X X X|
$$ | X X X X X X X X X X X X X X X X X X X|
$$ | X X X X X X X X X X X X X X X X X X .|
$$ | X X X X X X X X X X X X X X X X X X X|
$$ | X X X X X X X X X X X X X X X X X X C|
$$ | X X X X X X X X X X X X X X X X X O B|
$$ | X X X X X X X X X X X X X X X X X O O|
$$ | X X X X X X X X X X X X X X X X O O .|
$$ | X X X X X X X X X X X X X X X X O . O|
$$ ----------------------------------------[/go]
In hypothetical play White can take :bc: and fill the ko, so the :bc: stone is dead. That means that the :ec: 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. :lol: )

Re: Martian problem

Posted: Sun Sep 13, 2020 7:27 pm
by Matti
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.
You didn't state this in the problem.
Click Here To Show Diagram Code
[go]$$B
$$ -----------------------
$$ | . X O . O O X . X O |
$$ | X X O c . O X . X O |
$$ | . X X O O O X X O O |
$$ | X X O . O X X O . O |
$$ | O O O . O X W . O O |
$$ | X X O O O X X O O X |
$$ | . X X X O O O X X X |
$$ | X X X X X b O X . X |
$$ | X . . . X X O O X . |
$$ | X X . X X a X O X X |
$$ -----------------------[/go]

Re: Martian problem

Posted: Sun Sep 13, 2020 7:40 pm
by Matti
With japanese rules in hypothetical play the circled white stones can also be calied dead.
Click Here To Show Diagram Code
[go]$$B
$$ -----------------------
$$ | . X O . O O X W 3 O |
$$ | X X O O O O X W 1 O |
$$ | . X X O O O X X O O |
$$ | X X O O O X X 2 B O |
$$ | O O O . O X 4 B O O |
$$ | X X O O O X X O O X |
$$ | . X X X O O O X X X |
$$ | X X X X X O O X . X |
$$ | X . X . X X O O X . |
$$ | X X . X X X X O X X |
$$ -----------------------[/go]
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.

Re: Martian problem

Posted: Sun Sep 13, 2020 8:34 pm
by jaeup
Gérard TAILLE wrote:Hi Bill,

I am not sure to understand the japonese rule !
Click Here To Show Diagram Code
[go]$$B
$$ ----------------------------------------
$$ | X X X X X X X X X X X X X X X X X X X|
$$ | X X X X X X X X X X X X X X X X X X X|
$$ | X X X X X X X X X X X X X X X X X X X|
$$ | X X X X X X X X X X X X X X X X X X X|
$$ | X X X X X X X X X X X X X X X X X X X|
$$ | X X X X X X X X X X X X X X X X X X X|
$$ | X X X X X X X X X X X X X X X X X X X|
$$ | X X X X X X X X X X X X X X X X X X X|
$$ | X X X X X X X X X X X X X X X X X X X|
$$ | X X X X X X X X X X X X X X X X X X X|
$$ | X X X X X X X X X X X X X X X X X X X|
$$ | X X X X X X X X X X X X X X X X X X X|
$$ | X X X X X X X X X X X X X X X X X X .|
$$ | X X X X X X X X X X X X X X X X X X X|
$$ | X X X X X X X X X X X X X X X X X X O|
$$ | X X X X X X X X X X X X X X X X X O 1|
$$ | X X X X X X X X X X X X X X X X X O O|
$$ | X X X X X X X X X X X X X X X X O O .|
$$ | X X X X X X X X X X X X X X X X O . O|
$$ ----------------------------------------[/go]
In the position above you can see 349 black stones against 8 white stones and I suppose 341 white stones were killed.
After black :b1:, 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 ?
The simple answer is "yes". The reasoning is somewhat subtle. White may try two things.

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.
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