Life In 19x19

Bill Spight wrote:Let's suppose that you have a position where you can move to a local count of G with gote or S with sente

Hi, the discussion seems very interesting. Could you explain what are G and S, so that I can follow it from the beginning ?

Are you talking about the "swing values" of "local sequences" ? That is the local difference of points, after the frontiers are closed in the local area considered, between the case "Black plays first" and the case "White plays first" ?
This being counted with Territory scoring (area scoring give different swing values for gote sequences, because the number of Black and White stones is not equal. It also gives different values if the sequence is sente, but has optional follow-ups resulting in the same player playing locally twice in a row).

Also, when you say "can move to a local count of G with gote or S with sente", do you mean that the player has the choice between two endgame sequences, one gote with a swing value of G points, and another one sente with a swing value of S points ?

A local count is Black's points minus White's points.

I think your quotation refers to a local endgame in which one player chooses, on his first local move, between a gote option resulting in the position G with the local count C(G) or a different sente option resulting, if played as a local sente sequence, in the position S with the local count C(S).

We often use symbols conveniently with multiple meanings. E.g., we might use G and S also as the counts of these positions, i.e., C(G) = G and C(S) = S.

EDITS

Pio2001 wrote:

Bill Spight wrote:Let's suppose that you have a position where you can move to a local count of G with gote or S with sente

Hi, the discussion seems very interesting. Could you explain what are G and S, so that I can follow it from the beginning ?

Hi, Pio. Glad you find this interesting.

In the original discussion about RBerenguel's question, G and S are local counts, either scores or average scores before the position has been played out ( http://senseis.xmp.net/?Count ). In our current discussion I use them as local scores, and they are non-negative numbers.

Are you talking about the "swing values" of "local sequences" ?

In both the original discussion and this one we are comparing lines of play with the same player (Black, by default) playing first. Swing values would arise from comparisons with results when different players play first. In the first discussion I also talk about how much a play gains, which is only its swing value in the case of reverse sente. In our current discussion swing values may appear, such as in the comparison, 2B >= 2D. But other values appear, such as C + 2D, which are not swing values. These values are simply local results or differences in local results in different lines of play.

Edit: There are also swing values that do not appear in the current discussion. For instance, if {2B | 0 || -A} is gote, its swing value is A + B, but that value does not appear in any comparison. This discussion is not about swing values.

Also, when you say "can move to a local count of G with gote or S with sente", do you mean that the player has the choice between two endgame sequences, one gote with a swing value of G points, and another one sente with a swing value of S points ?

Berenguel was talking about local positions which offer a choice of sente or gote plays, where playing one option destroys the other. G and S in that sentence are the results of each choice, not the values of the moves. You can also have comparisons between independent gote and sente, but that is not what his original question was about.

RobertJasiek wrote:We often use symbols conveniently with multiple meanings. E.g., we might use G and S also as the counts of these positions, i.e., C(G) = G and C(S) = S.

It is my preference to use lower case letters for numbers and uppercase letters for positions and games. But lower case letters do not stand out well in the midst of text. So in these discussions I often use uppercase for both.

Click Here To Show Diagram Code

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

During the early endgame, which is Black's correct move?

Click Here To Show Diagram Code

[go]$$B Count +5
$$ -------------------
$$ | O . 1 . . . . X .
$$ | O X . X X . X X X
$$ | O X . O O X X X .
$$ | O O O O O O O X X
$$ | . O . O O O O O O[/go]

Click Here To Show Diagram Code

[go]$$B Count?
$$ -------------------
$$ | O . . . . . . X .
$$ | O X 1 X X . X X X
$$ | O X . O O X X X .
$$ | O O O O O O O X X
$$ | . O . O O O O O O[/go]

How to calculate the count? Is it also 5? I think that yes but I am not used to microendgame calculations. Is the next follow-up move a sente or reverse sente?

RobertJasiek wrote:

$$B Count?
$$ -------------------
$$ | O . . . . . . X .
$$ | O X 1 X X . X X X
$$ | O X . O O X X X .
$$ | O O O O O O O X X
$$ | . O . O O O O O O

Click Here To Show Diagram Code

[go]$$B Count?
$$ -------------------
$$ | O . . . . . . X .
$$ | O X 1 X X . X X X
$$ | O X . O O X X X .
$$ | O O O O O O O X X
$$ | . O . O O O O O O[/go]

How to calculate the count? Is it also 5? I think that yes but I am not used to microendgame calculations. Is the next follow-up move a sente or reverse sente?

I'm no endgame expert, but here's my take on the inquiry above.

From this position, it's gote both for white and black to play at 'a':

Click Here To Show Diagram Code

[go]$$B
$$ -------------------
$$ | O a . . . . . X .
$$ | O X X X X . X X X
$$ | O X . O O X X X .
$$ | O O O O O O O X X
$$ | . O . O O O O O O[/go]

Therefore, it's a 50% chance that black plays here and ends the position:

Click Here To Show Diagram Code

[go]$$B Position A
$$ -------------------
$$ | O B . . . . . X .
$$ | O X X X X . X X X
$$ | O X . O O X X X .
$$ | O O O O O O O X X
$$ | . O . O O O O O O[/go]

In this case, that's 6 points for black with 50% probability. What about the other 50% chance? Well, that comes down to the count of this position, where white played first:

Click Here To Show Diagram Code

[go]$$B Position B
$$ -------------------
$$ | O W . . . . . X .
$$ | O X X X X . X X X
$$ | O X . O O X X X .
$$ | O O O O O O O X X
$$ | . O . O O O O O O[/go]

So what's this count? Recursively apply the procedure above... GIVEN this position, there's 50% chance that black blocks:

Click Here To Show Diagram Code

[go]$$B Position C
$$ -------------------
$$ | O O B . . . . X .
$$ | O X X X X . X X X
$$ | O X . O O X X X .
$$ | O O O O O O O X X
$$ | . O . O O O O O O[/go]

So 50% chance of getting 5 points. But what about the other 50%? Well, it's the value of this position:

Click Here To Show Diagram Code

[go]$$B Position D
$$ -------------------
$$ | O O W . . . . X .
$$ | O X X X X . X X X
$$ | O X . O O X X X .
$$ | O O O O O O O X X
$$ | . O . O O O O O O[/go]

Repeat procedure, again. 50% chance of black playing first and ending, this time for 4 points:

Click Here To Show Diagram Code

[go]$$B Position E
$$ -------------------
$$ | O O O B . . . X .
$$ | O X X X X . X X X
$$ | O X . O O X X X .
$$ | O O O O O O O X X
$$ | . O . O O O O O O[/go]

And 50% chance that it's the value of this position:

Click Here To Show Diagram Code

[go]$$B Position F
$$ -------------------
$$ | O O O W . . . X .
$$ | O X X X X . X X X
$$ | O X . O O X X X .
$$ | O O O O O O O X X
$$ | . O . O O O O O O[/go]

which is 50% chance of this:

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[go]$$B Position G
$$ -------------------
$$ | O O O O B . . X .
$$ | O X X X X . X X X
$$ | O X . O O X X X .
$$ | O O O O O O O X X
$$ | . O . O O O O O O[/go]

(3 points) vs. 50% chance of this (1 point):

Click Here To Show Diagram Code

[go]$$B Position H
$$ -------------------
$$ | O O O O W . . X .
$$ | O X X X X . X X X
$$ | O X . O O X X X .
$$ | O O O O O O O X X
$$ | . O . O O O O O O[/go]

At this point, it's questionable as to whether the white move is gote. Black gets a point by responding, so it's relatively large. So I'd say that it's likely going to elicit a response from black. In other words, the board position will end like this, given the scenario above:

Click Here To Show Diagram Code

[go]$$B Position I
$$ -------------------
$$ | O O O O O B . X .
$$ | O X X X X . X X X
$$ | O X . O O X X X .
$$ | O O O O O O O X X
$$ | . O . O O O O O O[/go]

for a final position of 1 points. That's a terminal state.

Now roll the evaluation back up to the beginning.

0.5*Value(A)+0.5*(0.5*Value(C)+0.5*(0.5*Value(E)+0.5*(0.5*Value(G)+0.5*Value(H))))

Which comes to:

0.5*6+0.5(0.5*5+0.5*(0.5*4+0.5*(0.5*3+0.5*1))) = 5 points...

That's my interpretation, anyway, but in a real game, I'd say, something like, "meh, maybe a bit over 5 points?".

How far off am I in this calculation?

Yeah, the count is clearly exactly 5(and all moves are worth 1 point exactly. But I think this second Position favors black compared to the first one, since he can kind of decide, when to block, which will be useful for tedomari. Bill can explain this nicely with his infinitesimals(should be a corridor ending in a tiny?)

I mean he can decide because the white pushes are ambigous between sente and gote, they can be treated as 1pt sente Or 2pt gote, which is the same size, and this decision is Up to black.

Sorry For some uppercase words, hard to Write an english Text versus your German Autocorrect.

It's not "clearly" 5 to me, unless there's a method I'm not aware of. For example, if we modify the position slightly such that the end position in my analysis were two points, as follows:

Click Here To Show Diagram Code

[go]$$B
$$ -------------------
$$ | O . . . . . . X .
$$ | O X X X X . X X X
$$ | O X . O B X X X .
$$ | O O O O O O O X X
$$ | . O . O O O O O O[/go]

I calculate a value of 5.0625...

If that's correct, it's not clear to me unless I do that long calculation..

My "rule" is: push in a corridor that ends with sente or with 2 point gote always has a miai value of 1, no matter how long the corridor.

Actually, according to this rule, you should be mistaken.., this ends in 2 point gote, so the miai value should still be 5(that stone beeing black does help black anyway though , since its even better for tedomari.

Actually, there was a mistake in your analysis before: If you treat the last white move as sente, it has Not only 50% Chance of Happening. But this mistake cancelled out, because you should have treated one push earlier(Position f) as sente already, because the withe Stone Makes that last push big and black should prevent it(F always leads to G if black plays correctly, I does Not happen) But you can see how that first mistake infuenced the other count in the modified Position , where the end is 3 and Not 0,5(3+2).

Let me try again:

Position D is {4 ||| 3 || 1 || 0 | BIG} = {4 || 3 | 0} because the right subtree is White's sente. {3 | 0} has the count 1.5 and the move value 1.5.

If we assumed the root to be gote, we would be calculating its tentative count from {4 | 1.5} as 2.75 and its tentative move value as 1.25. Since this is smaller than the follow-up move value 1.5, actually the root is sente for White. Position D has the inherited count 3 from the sente follower.

Therefore, the position B is {5 | 3} = 4.

Therefore, the initial position (black stone at the 3-2 point) is {6 | 4} = 5.

I guess, there might also be some infinitesemal but which and why.

Hi, guys!

Click Here To Show Diagram Code

[go]$$W Position D
$$ -------------------
$$ | O O O 1 2 . . X . |
$$ | O X X X X a X X X |
$$ | O X . O O X X X . |
$$ | O O O O O O O X X |
$$ | . O . O O O O O O |
$$ -------------------[/go]

As Schachus and Robert indicate, is sente. That being the case the count of position D is 3.

You can tell that it is sente because if White gets in a play at 2 it takes away the point at "a", which does not do. The position after actually has a count of 0, as Robert says, but you don't have to do the calculation to see that is sente.

Click Here To Show Diagram Code

[go]$$B
$$ -------------------
$$ | O . . . . . B X . |
$$ | O X X X X B . . X |
$$ | O X . O O X X X . |
$$ | O O O O O O O X X |
$$ | . O . O O O O O O |
$$ -------------------[/go]

This position has a count of 5.0625.

More in the morning.

I'm not sure, how much better the solid connection really is for black, but here another argument(whitout any count or infinitesimals), why it can only be better, never worse than the first line move:

Click Here To Show Diagram Code

[go]$$B
$$ -------------------
$$ | O a 1 . . . . X .
$$ | O X . X X . X X X
$$ | O X b O O X X X .
$$ | O O O O O O O X X
$$ | . O . O O O O O O[/go]

This position is finished, white has a ko threat at either a or b. Lets assume he is going to use a later(it doesnt make a difference, but it makes the comparison easier. Edit: actually, a is the bigger threat anyway). Hence locally play will go like this

Click Here To Show Diagram Code

[go]$$B
$$ -------------------
$$ | O 2 1 . . . . X .
$$ | O X 3 X X . X X X
$$ | O X . O O X X X .
$$ | O O O O O O O X X
$$ | . O . O O O O O O[/go]

Now in the other Variation, the only local move for white is also going to be at 2:

Click Here To Show Diagram Code

[go]$$B
$$ -------------------
$$ | O 2 3 . . . . X .
$$ | O X 1 X X . X X X
$$ | O X . O O X X X .
$$ | O O O O O O O X X
$$ | . O . O O O O O O[/go]

. But here black has the choice, whether to block 3 leading to exactly the same result(!, so black can make SURE, he is not worse off then in the other line), or whether not to block. And since 3 has the same value(miai speeking) as 2, this option not to answer is not really a big benefit, but its also not a stpid option you would never use. Also since 2 is not strictly stente for white, also blaxk might get to play there, which is extra option the other line doesnt have. Last but not least, like this there is no ko threat for w.

I have not mentioned the "no kos now or later" condition but meant it because, if you really want to discuss ko threats (or kos created locally...!), it is a never-ending topic and iterative ko threats (becoming effective after a succession of such plays of a player) would also be relevant. If Black chooses the solid connection, White can play 5 successive plays to have a ko threat.

Bill Spight wrote:As Schachus and Robert indicate, is sente. That being the case the count of position D is 3.

Ok. Then I don't understand why the count of the original position is not this:
0.5*6+0.5(0.5*5+0.5*(0.5*4+0.5*(3))) = 5.125.

My calculations are from these positions:

Click Here To Show Diagram Code

[go]$$B Pos A
$$ -------------------
$$ | O B . . . . . X .
$$ | O X X X X . X X X
$$ | O X . O O X X X .
$$ | O O O O O O O X X
$$ | . O . O O O O O O[/go]

Click Here To Show Diagram Code

[go]$$B Pos B
$$ -------------------
$$ | O W . . . . . X .
$$ | O X X X X . X X X
$$ | O X . O O X X X .
$$ | O O O O O O O X X
$$ | . O . O O O O O O[/go]

Click Here To Show Diagram Code

[go]$$B Pos C
$$ -------------------
$$ | O O B . . . . X .
$$ | O X X X X . X X X
$$ | O X . O O X X X .
$$ | O O O O O O O X X
$$ | . O . O O O O O O[/go]

Click Here To Show Diagram Code

[go]$$B Pos D
$$ -------------------
$$ | O O W . . . . X .
$$ | O X X X X . X X X
$$ | O X . O O X X X .
$$ | O O O O O O O X X
$$ | . O . O O O O O O[/go]

Click Here To Show Diagram Code

[go]$$B Pos E
$$ -------------------
$$ | O O O B . . . X .
$$ | O X X X X . X X X
$$ | O X . O O X X X .
$$ | O O O O O O O X X
$$ | . O . O O O O O O[/go]

Click Here To Show Diagram Code

[go]$$B Pos F (since white's push is sente)
$$ -------------------
$$ | O O O W B . . X .
$$ | O X X X X . X X X
$$ | O X . O O X X X .
$$ | O O O O O O O X X
$$ | . O . O O O O O O[/go]

My calculation is as follows:

From the initial position, both black and white plays are gote. So there's 50% chance of either.

The count is:
Initial Count = 0.5*Count(Pos A) + 0.5*Count(Pos B)

The count of Pos A is clear since it's terminal (6 points), but the count of Pos B is:
Count(Pos B) = 0.5*Count(Pos C) + 0.5*Count(Pos D)

The count of Pos C is clear since it's terminal (5 points), but the count of Pos D is:
Count(Pos D) = 0.5*Count(Pos E) + 0.5*Count(Pos F)

Both are terminal since position F results in a black response (since it's sente), and the values are 4 and 3, respectively.

Therefore, the initial count is:
Initial Count = 0.5*6+0.5(0.5*5+0.5*(0.5*4+0.5*3)) = 5.125.

What's wrong with this approach to the count?

Kirby wrote:

Bill Spight wrote:As Schachus and Robert indicate, is sente. That being the case the count of position D is 3.

My calculations are from these positions:
[...]

$$B Pos D
$$ -------------------
$$ | O O W . . . . X .
$$ | O X X X X . X X X
$$ | O X . O O X X X .
$$ | O O O O O O O X X
$$ | . O . O O O O O O

Click Here To Show Diagram Code

[go]$$B Pos D
$$ -------------------
$$ | O O W . . . . X .
$$ | O X X X X . X X X
$$ | O X . O O X X X .
$$ | O O O O O O O X X
$$ | . O . O O O O O O[/go]

$$B Pos E
$$ -------------------
$$ | O O O B . . . X .
$$ | O X X X X . X X X
$$ | O X . O O X X X .
$$ | O O O O O O O X X
$$ | . O . O O O O O O

Click Here To Show Diagram Code

[go]$$B Pos E
$$ -------------------
$$ | O O O B . . . X .
$$ | O X X X X . X X X
$$ | O X . O O X X X .
$$ | O O O O O O O X X
$$ | . O . O O O O O O[/go]

$$B Pos F (since white's push is sente)
$$ -------------------
$$ | O O O W B . . X .
$$ | O X X X X . X X X
$$ | O X . O O X X X .
$$ | O O O O O O O X X
$$ | . O . O O O O O O

Click Here To Show Diagram Code

[go]$$B Pos F (since white's push is sente)
$$ -------------------
$$ | O O O W B . . X .
$$ | O X X X X . X X X
$$ | O X . O O X X X .
$$ | O O O O O O O X X
$$ | . O . O O O O O O[/go]

My calculation is as follows:
[...]
The count of Pos C is clear since it's terminal (5 points), but the count of Pos D is:
Count(Pos D) = 0.5*Count(Pos E) + 0.5*Count(Pos F)



What's wrong with this approach to the count?

Everything until here is right. But the count of position D is the same as the count of position F("sente gains nothing"). Position E is only important if you want to previous white push (the one leading from B to D) as sente, if you count that move gote, that means black will not answer and thus white get the next push as his privilege because its sente. Hence count of D is the count of F. You can only average the counts if play is gote for BOTH sides.