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AmbiguousPosition

 

Ambiguous Position Discussion
Path: CGTPath   · Prev: InfinitesimalsDiscussion   · Next: Chilling
   

[Diagram]
Diag.: Ambiguous position

Black has a one-point gote play (miai counting) with Black a - White b, Black c, or a one-point sente play with Black c - White a.



Not sure if I can follow your counting here. If Black plays c, then White will get the additional point at b, since in the given position, there's no need to defend further. Consequently, if the gote move is considered to be worth one point for Black, then the sente move is worth zero (ignoring the reverse gote possibility for White, of course). --Georg Mischler 2k

Let's normalize the count, so that the result if Black plays c and White plays a is 0. Then we may represent the position like this:

          {{Big | 0}, 1 | -1}

("Big" means big enough for sente. Here it is more than 5, I think. Negative numbers are White's scores. Positions or scores to the left of the bar are reached by Black's moves, those to the right are reached by White's moves.)

          {1 | -1}

is a one-point gote play.

           {{Big | 0} | -1}

is a one-point sente plaY.

Make sense?

-- BillSpight

Assuming that I understand your notation correctly: If the {{Big | 0} | -1} is meant to be one-point sente (0 - -1 = 1), then {1 | -1} still looks a two-point sente to me in comparison (1 - -1 = 2). Please don't get me wrong, I really like the discussion about chosing the "right" move here, especially considering the (still sente!) "Big" follow-up. I just don't see the ambiguity. Or maybe my real world understanding of the term "ambiguity" is too narrow to be applied here. Now if we could determine the exact point value of getting sente, things might become much more obvious... --Georg Mischler

Sorry. I have amended the original to state that I am using miai counting. That's the one to use when comparing plays. You are using deiri counting, which is more common. The rule for comparing deiri plays is to multiply sente values by 2 (or, equivalently, to divide gote values by 2, which gives you miai values). So the two-point gote and the one-point sente are the same size.

-- Bill

Ah well, even I can understand it if you explain it this way! ;-) --GeorgMischler
(PS: Maybe this difference between the value of sente and gote moves should be explicitly mentioned in MiaiCounting then?)


--Jesusin (3K): I am confused about the following position.

[Diagram]
Diag.: Black to move

The play at a is White's sente. Black's move at a is worth one point in reverse sente. At f is worth one point in gote (MiaiCounting). So both moves should be equivalent, but...

1. a b c d e f leads to a jigo;

2. f g d a c leads to Black winning by one point.

Please help me! I can't understand it!



Bill: First, to address the life and death issue.

[Diagram]
Diag.: Ko

After White 2, Black 3 makes ko. If White 4, Black 5 and White cannot play at a.


[Diagram]
Diag.: Ko (ii)

Another way to make ko.


[Diagram]
Diag.: No ko


Let's make it easier by giving White another eye. :-)

[Diagram]
Diag.: Which one-point play?


Here a is a simple one-point gote play. At b is an ambiguous one-point play. They both have the same value, but Black to play prefers a, while White to play prefers b. After White b, a and c are miai.

The players prefer different plays because they are trying to get tedomari, the last play. Getting the last (one point) play does make a difference of one point, here.

Getting the last one-point play is discussed in Chilling, Infinitesimals, Playing Infinitesimals, and so on (CGT path).

In terms of infinitesmals a is * (STAR) and b is ^ (UP). ^ has an atomic weight of 1, while * has an atomic weight of 0. Atomic weight is similar to the count of external liberties in a semeai. A white play at b changes the local position (c) to *, with atomic weight of 0, reducing Black's atomic weight by 1. That is like filling one of Black's liberties in a semeai. A Black play at a does not change the atomic weight. That is like not filling one's own liberty.

[Diagram]
Diag.: Which one-point play? (ii)

But you have to take miai into account. In this diagram a and d are miai. That means that Black should play at b to get tedomari.


[Diagram]
Diag.: Which play? (ii) Correct

Black wins by one point.


[Diagram]
Diag.: Which play? (ii) Error

If Black 1 (or 6), White gets tedomari, for jigo.

--BillSpight


--Jesusin (3K): SL's 'LordOfTheYose' is always right ;-). Thanks for the enlightment, Bill.

I have been working on this and I now think b in 'Diag:Which one-point play?' is sente for White and is a 1/2 point gote play for Black. I will try to prove it:

[Diagram]
Diag.: Black to move

Here a is 1/2 point gote. If it was White#s turn, she would take her sente at b and then a to win. If it is Black's turn, Black b, White plays atari, Black connects, White a is jigo. And Black a, White b, Black connects is jigo too.



Bill: Basically the size of a play is a local phenomenon. It is not true that, on the whole board, the largest play is always the best play, nor that two plays of different sizes always lead to different results.
Let's alter your example slightly.

[Diagram]
Diag.: White to move

White a has a miai value of 1/2, while White b has a miai value of 3/4. On this board, however, White gets the same result with either.


[Diagram]
Diag.: White plays 1/2 point play

After White 1, White 3 gets tedomari, for jigo.


[Diagram]
Diag.: White plays 3/4 point play

Now Black gets tedomari, but the result is still jigo.

In the first case, White gained 1/2 point, then lost 3/4 point, then gained 1/2 point, for a net gain of 1/4 point. In the second case, White gained 3/4 point and then lost 1/2 point, for a net gain of 1/4 point. All same same.

In your example if Black plays the 1/2 point play first, he gains 1/2 point, as White's sente sequence gains no points. If Black plays the one-point play first, he gains one point but then loses 1/2 point. That's why the result is the same. :-)



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