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 Post subject: Senseis Endgame Problem Riddle
Post #1 Posted: Sat Aug 02, 2014 11:01 am 
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Hi,

the following endgame problem is from senseis and I would really appreciate if someone could explain me how to make/draw a count tree and temperature tree for the plays around C.
A and B i figuered out with my mumbo-jumbo system.
Thanks (To the author Bill, great problem!)

Click Here To Show Diagram Code
[go]$$B Tedomari problem 1
$$ -------------------
$$ | . . X X a X O . . |
$$ | X X . X O . O . . |
$$ | X O X X O O O O O |
$$ | X O O O . O X X b |
$$ | O O . O O O O O X |
$$ | . O O X O X O X X |
$$ | . O X X X X X . X |
$$ | O O c O X . . X . |
$$ | . X . X X . . . . |
$$ -------------------[/go]


And heres the Link to the page: Endgame Problem 24

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 Post subject: Re: Senseis Endgame Problem Riddle
Post #2 Posted: Sat Aug 02, 2014 5:23 pm 
Gosei
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Starting from this position:

Click Here To Show Diagram Code
[go]$$B Tedomari problem 1
$$ -------------------
$$ | . . X X a X O . . |
$$ | X X . X O . O . . |
$$ | X O X X O O O O O |
$$ | X O O O . O X X b |
$$ | O O . O O O O O X |
$$ | . O O X O X O X X |
$$ | . O X X X X X . X |
$$ | O O O O X . . X . |
$$ | . X c X X . . . . |
$$ -------------------[/go]

If white plays c, the score is -3 (white points are treated as negative). If black plays c, the score is 0. So the current score in the bottom left is the sum of the scores divided by two, which is -3/2. A play by either player is worth the absolute value of the difference of the scores divided by two, which is also 3/2.

Click Here To Show Diagram Code
[go]$$B Tedomari problem 1
$$ -------------------
$$ | . . X X a X O . . |
$$ | X X . X O . O . . |
$$ | X O X X O O O O O |
$$ | X O O O . O X X b |
$$ | O O . O O O O O X |
$$ | . O O X O X O X X |
$$ | . O X X X X X . X |
$$ | O O c O X . . X . |
$$ | . X . X X . . . . |
$$ -------------------[/go]

From this position, if white plays c, we know from the result above that the score will be -3/2. If black plays c, the score is 2. The current score in the area is the sum of the scores divided by two: 1/4. A play by either side is worth the absolute value of the difference of the scores divided by two: 7/4.

So, c is worth more than a but less than b. But because of tedomari, with black to move first black should play c, not b.

I've done these calculations in my head before, but never actually made the steps explicit like this. I think the steps are correct but I could be proven wrong :).

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 Post subject: Re: Senseis Endgame Problem Riddle
Post #3 Posted: Sun Aug 03, 2014 11:19 am 
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Thank you for the answer.
I also got the miai value of 1 3/4 for "c" and a count of 1/4; on the sensei page there is a count of - 1 1/2 and i dont get that part.

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 Post subject: Re: Senseis Endgame Problem Riddle
Post #4 Posted: Sun Aug 03, 2014 12:03 pm 
Honinbo

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Xylol wrote:
Thank you for the answer.
I also got the miai value of 1 3/4 for "c" and a count of 1/4; on the sensei page there is a count of - 1 1/2 and i dont get that part.


1/4 - 1 3/4 = -1 1/2

Local count after a White move.

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