A special (new?) strategy in AGA field?

[jmeinh]

Is that really a problem?

I think it's enough for Black to make the appropriate defence moves when necessary; then he continues the interrupted loop, and W can't defend against that.
Or am I missing something?

[Gérard TAILLE]

Is that really a problem?

I think it's enough for Black to make the appropriate defence moves when necessary; then he continues the interrupted loop, and W can't defend against that.
Or am I missing something?

Yes it is a problem

The problem is due to the subtilities of SSK : at the moment black makes the appropiate defence move then white passes and after this pass the loop can no longer work!

[jmeinh]

I see. Then my idea doesn't work as intended.

[hzamir]

Black to play and win

I must be missing something, what is the number of prisoners on each side?
At first sight, assumming no prisoners, black has 22 points, white has 14, so with komi 6.5 black won already.

While not especially necessary, AGA still counts with prisoners as a convention.
Black to play, means an equal number of moves have been made.
43 White stones on the board
38 Black stones on the board
White has thus captured 5 more prisoners than black. Even in unlike event of an early pass in the game, player would still have to hand over a prisoner in AGA.

[Gérard TAILLE]

Black to play and win

I must be missing something, what is the number of prisoners on each side?
At first sight, assumming no prisoners, black has 22 points, white has 14, so with komi 6.5 black won already.

While not especially necessary, AGA still counts with prisoners as a convention.
Black to play, means an equal number of moves have been made.
43 White stones on the board
38 Black stones on the board
White has thus captured 5 more prisoners than black. Even in unlike event of an early pass in the game, player would still have to hand over a prisoner in AGA.

Yes hzamir, providing you count 5 more prisoners for white you can count by territory the initial position:

$$B Black to play and win
$$ ------------------------
$$| T O T O T O T O T O T |
$$| O O O O O O O O O O Y |
$$| X X X X X X O T O T T |
$$| M M M M M X X O T O T |
$$| M M M M M M X O O O O |
$$| M X M M M M X X X X X |
$$| X X X M M M X O O O O |
$$| O O X M M M X O T O T |
$$| . O X X X X X O O X O |
$$| X O O O O X O O X X X |
$$| M X X . O X O X M X M |
$$ -----------------------

Click Here To Show Diagram Code

[go]$$B Black to play and win
$$ ------------------------
$$| T O T O T O T O T O T |
$$| O O O O O O O O O O Y |
$$| X X X X X X O T O T T |
$$| M M M M M X X O T O T |
$$| M M M M M M X O O O O |
$$| M X M M M M X X X X X |
$$| X X X M M M X O O O O |
$$| O O X M M M X O T O T |
$$| . O X X X X X O O X O |
$$| X O O O O X O O X X X |
$$| M X X . O X O X M X M |
$$ -----------------------[/go]

black : 25 territory = 25 points
white : 14 territory + 1 prisoner (upper right corner) + 5 prisoners (to reach the initial position) = 20 points
With a komi of 6.5 black loses by 1.5 points (same result under https://lifein19x19.com/viewtopic.php?p=279412#p279412)

[Gérard TAILLE]

I see. Then my idea doesn't work as intended.

Anyway your idea is a major part of the solution but OC some more work is needed to really find it.

[Gérard TAILLE]

Ten days ago I proposed in https://lifein19x19.com/viewtopic.php?p=279399#p279399 a rather difficult problem with a superko and a large black territory not so easy to handle.
Maybe it's time to help you a little and I can propose the following simplier position on a small 9x9 board.

AGA rules. No komi

$$B Black to play
$$ ------------------------
$$| . X X . O O X . . |
$$| X O O O O X X X O |
$$| . O X X X . . X . |
$$| O O X X X X X X X |
$$| X X X O X O O O O |
$$| O O O O X O . O X |
$$| . O O X X O O X . |
$$| O O O X O O X X X |
$$| . . O X O . O X . |
$$ -----------------------

Click Here To Show Diagram Code

[go]$$B Black to play
$$ ------------------------
$$| . X X . O O X . . |
$$| X O O O O X X X O |
$$| . O X X X . . X . |
$$| O O X X X X X X X |
$$| X X X O X O O O O |
$$| O O O O X O . O X |
$$| . O O X X O O X . |
$$| O O O X O O X X X |
$$| . . O X O . O X . |
$$ -----------------------[/go]

In this position you can count 33 black stones and 33 white stones.
What is the best sequence to reach the best result for both?

To my point of view the solution is still not so easy to find and the resulting position is extremely surprising. I like very much this problem.
Because all subtilities in this problem are present in my previous problem it is certainly a good idea to firstly find the solution of this problem before resolving my previous one.

[Gérard TAILLE]

Just to be clear:

$$B :w2: :w4: :w6: :b7: pass
$$ ------------------------
$$| 1 X X . O O X . 5 |
$$| X O O O O X X X O |
$$| . O X X X . . X 3 |
$$| O O X X X X X X X |
$$| X X X O X O O O O |
$$| O O O O X O . O X |
$$| . O O X X O O X . |
$$| O O O X O O X X X |
$$| . . O X O . O X . |
$$ -----------------------

Click Here To Show Diagram Code

[go]$$B :w2: :w4: :w6: :b7: pass
$$ ------------------------
$$| 1 X X . O O X . 5 |
$$| X O O O O X X X O |
$$| . O X X X . . X 3 |
$$| O O X X X X X X X |
$$| X X X O X O O O O |
$$| O O O O X O . O X |
$$| . O O X X O O X . |
$$| O O O X O O X X X |
$$| . . O X O . O X . |
$$ -----------------------[/go]

B+5 but it is not satisfactory for black. You have to find something better.