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 Post subject: How superko can change a losing position into a winning one
Post #1 Posted: Tue Sep 01, 2020 1:30 pm 
Lives with ko

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Black to play and win
Click Here To Show Diagram Code
[go]$$B
$$ -----------
$$ | O . a . O -
$$ | O O O O O -
$$ | O O O O O -
$$ | O O O O O -
$$ | O O O O O -
$$ -----------[/go]


I suppose we are using AGA rule (SSK) on a 5x5 board.

The problem I propose is the following:

Starting with an empty board, what is the minimum number of moves you need to build the above position such that black can play and win.

Note : if for example you play 44 moves (22 pass for black + 22 white stones played by white) then the above position is a losing one for black because after a black move on "a" the position looks like a strange seki with only one black stone involved! Is this is true then it is a large win for white.

You have to find something else and minimise the number of moves!

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 Post subject: Re: How superko can change a losing position into a winning
Post #2 Posted: Tue Sep 01, 2020 3:43 pm 
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Why would that be a seki?

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 Post subject: Re: How superko can change a losing position into a winning
Post #3 Posted: Wed Sep 02, 2020 1:48 am 
Lives with ko

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Harleqin wrote:
Why would that be a seki?


The point is the following: after the sequence

Click Here To Show Diagram Code
[go]$$B
$$ -----------
$$ | O 3 1 4 O -
$$ | O O O O O -
$$ | O O O O O -
$$ | O O O O O -
$$ | O O O O O -
$$ -----------[/go]

:w2: pass
:b5: on :b1:
:w6: on :b3:
:b7: on :b1:

it follows

Click Here To Show Diagram Code
[go]$$Wm8
$$ -----------
$$ | . . X . . -
$$ | . . . . . -
$$ | . . 1 . . -
$$ | . . . . . -
$$ | . . . . . -
$$ -----------[/go]


and here my impression is that black cannot live somewhere on the board. Maybe I am not strong enough to show this and I hope stronger player will be able to confirm my impression.

If it is true the position is seki because after

Click Here To Show Diagram Code
[go]$$B
$$ -----------
$$ | O 2 1 3 O -
$$ | O O O O O -
$$ | O O O O O -
$$ | O O O O O -
$$ | O O O O O -
$$ -----------[/go]


we reach the position

Click Here To Show Diagram Code
[go]$$Wm4
$$ -----------
$$ | . . X X . -
$$ | . . . . . -
$$ | . . 1 . . -
$$ | . . . . . -
$$ | . . . . . -
$$ -----------[/go]


and now it appears to me that black can live with a better result than the initial position after black 1

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