Sente, gote and endgame plays

Kirby · Post by **Kirby** » Sat Aug 19, 2017 10:30 am

Ah, right. Position D is worth 3 since white will get the sente.

So Count(Pos B) = 0.5*Count(Pos C) + 0.5*3 = 2.5+1.5= 4

And the initial is
0.5*Count(Pos A) + 0.5*Count(Pos B) = 3+2=5, right?

Schachus · Post by **Schachus** » Sat Aug 19, 2017 10:38 am

Kirby wrote:Ah, right. Position D is worth 3 since white will get the sente.

So Count(Pos B) = 0.5*Count(Pos C) + 0.5*3 = 2.5+1.5= 4

And the initial is
0.5*Count(Pos A) + 0.5*Count(Pos B) = 3+2=5, right?

Right, thats what I get.

Bill Spight · Post by **Bill Spight** » Sat Aug 19, 2017 11:11 am

If the original position is worth 5.125, that means that 8 of them are worth 41. So we would normally expect that Black could get 41 points, even if White played first. Let's see.

Click Here To Show Diagram Code: [go]$$W Position B $$ ------------------- $$ | O 1 . . . . . X . | $$ | 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]$$W Position A $$ ------------------- $$ | O 2 . . . . . X . | $$ | 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]$$W Position B $$ ------------------- $$ | O 3 . . . . . X . | $$ | 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]$$W Position A $$ ------------------- $$ | O 4 . . . . . X . | $$ | 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]$$W Position B $$ ------------------- $$ | O 5 . . . . . X . | $$ | 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]$$W Position A $$ ------------------- $$ | O 6 . . . . . X . | $$ | 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]$$W Position B $$ ------------------- $$ | O 7 . . . . . X . | $$ | 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]$$W Position A $$ ------------------- $$ | O 8 . . . . . X . | $$ | 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]

If Black treats

as sente, she gets only 40 points, so she does not reply to it, or the other initial plays in each copy. After 8 moves Black has 24 points in four of the copies. Let's continue in the other four.

Click Here To Show Diagram Code: [go]$$Wm9 Position F $$ ------------------- $$ | O W 1 3 4 . . X . | $$ | 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]$$Wm9 Position C $$ ------------------- $$ | O W 2 . . . . X . | $$ | 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]$$Wm9 Position F $$ ------------------- $$ | O W 5 7 8 . . X . | $$ | 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]$$Wm9 Position C $$ ------------------- $$ | O W 6 . . . . X . | $$ | 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]

Again, if Black treated

as sente she would get only 40 points, so she does not answer it. However,

is sente, and Black answers it. So is

.

The end result is 24 + 10 + 6 = 40 points for Black. You may verify that if Black plays first the result is only 41 points for Black. The average result is 40.5, which shows that the average value (count) of the original position is less than 5.125. If we have 16 copies, White to play will hold Black to 80 points, while Black to play will get 81 points, which means that the count of the original position is less than 5.0625. Et cetera, et cetera. The count is at least 5 and less that 5 plus any fraction.

Schachus · Post by **Schachus** » Sat Aug 19, 2017 11:25 am

Am I right Bill, that this 1 point extra black in these many copies if he starts is because he always gets tedomari(that is his advantage here) and if he starts, tedomari means he gets one more 1pt move?!

Bill Spight · Post by **Bill Spight** » Sat Aug 19, 2017 11:39 am

Schachus wrote:Am I right Bill, that this 1 point extra black in these many copies if he starts is because he always gets tedomari(that is his advantage here) and if he starts, tedomari means he gets one more 1pt move?!

That's a good observation.

Bill Spight · Post by **Bill Spight** » Sat Aug 19, 2017 11:57 am

Comparison of original plays by Black

Click Here To Show Diagram Code: [go]$$W Kosumi $$ ------------------- $$ | O . B . . . . X . | $$ | 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]

The count is 5 exactly.

Click Here To Show Diagram Code: [go]$$W Solid connection $$ ------------------- $$ | O . . . . . . X . | $$ | O X B 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]

The count is also 5, but the method of multiples in my previous note indicates that Black to play can do better than average in any finite number of copies of this position. So we may consider the value of this position to be infinitesimally better than 5. (This is not what we mean by go infinitesimals, BTW. They are positions where a gote or reverse sente or ambiguous play gains one point. See https://senseis.xmp.net/?GoInfinitesimals .)

Or, more simply, as Schachus's observation about tedomari indicates, Black can guarantee a score of 5, even if White plays first, and White cannot guarantee a score of only 5, if Black plays first. So the value of this position (for Black) is greater than 5.

Bill Spight · Post by **Bill Spight** » Sat Aug 19, 2017 4:29 pm

Comparing plays by difference games

To get started, let’s set up a 0 game by reversing the colors of the stones in a mirror position.

Click Here To Show Diagram Code: [go]$$B $$ ------------------- $$ | O . . . . . . X . | $$ | 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]

Click Here To Show Diagram Code: [go]$$B $$ ------------------- $$ | X . . . . . . O . | $$ | X O . O O . O O O | $$ | X O . X X O O O . | $$ | X X X X X X X O O | $$ | . X . X X X X X X | $$ -------------------[/go]

First, let Black play the kosumi and then let White play the solid connection in the mirror position.

Click Here To Show Diagram Code: [go]$$B $$ ------------------- $$ | O . 1 . . . . X . | $$ | 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]$$B $$ ------------------- $$ | X 3 4 . . . . O . | $$ | X O 2 O O . O O O | $$ | X O . X X O O O . | $$ | X X X X X X X O O | $$ | . X . X X X X X X | $$ -------------------[/go]

The result is jigo. So White's play, the solid connection, is at least as good as Black's play, the kosumi. White can let Black play first and still get jigo.

Next, let Black play the solid connection and White play the kosumi.

Click Here To Show Diagram Code: [go]$$B $$ ------------------- $$ | O 3 . . . . . X . | $$ | 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]

Click Here To Show Diagram Code: [go]$$B $$ ------------------- $$ | X . 2 . . . . O . | $$ | X O . O O . O O O | $$ | X O . X X O O O . | $$ | X X X X X X X O O | $$ | . X . X X X X X X | $$ -------------------[/go]

Black wins by 1 point. Putting the two results together we conclude that the solid connection is superior to the kosumi, since it may win the difference game but never loses it.

It still may be that the kosumi is the way to play if there is a ko fight or a potential ko fight, but otherwise, the solid connection is better.

Kirby · Post by **Kirby** » Sat Aug 19, 2017 5:10 pm

Considering 8, 16, etc. copies of the same position is creative.

I can kind of see how the position discussed earlier was slightly above 5, but less than 5 and any fraction.

Are all such cases equal in count? For example, if I have two arbitrary shapes I count to be 5 plus an inexpressible fraction of a point, does it matter which I play first?

Schachus · Post by **Schachus** » Sat Aug 19, 2017 5:37 pm

Yes that does matter. You can have a look here viewtopic.php?f=15&t=14292
where Bill explained some of it to me

RobertJasiek · Post by **RobertJasiek** » Sat Aug 19, 2017 10:52 pm

The difference game I find much easier to understand than the multiples approach, which relies very much on hand-waving which sequences "obviously" need not be considered.

I still do not understand which infinitesimals describe the initial position of the solid connection, and how do we determine them?

Bill Spight · Post by **Bill Spight** » Sun Aug 20, 2017 1:34 am

RobertJasiek wrote:The difference game I find much easier to understand than the multiples approach, which relies very much on hand-waving which sequences "obviously" need not be considered.

With the method of multiples I was offering a demonstration, not a proof. You will find, if you have not already, that exhaustive search will produce the same conclusion. The method of multiples was used to prove the mean value theorem, back in the mid-20th century.

I still do not understand which infinitesimals describe the initial position of the solid connection, and how do we determine them?

Using CGT notation, here is the game tree (ignoring dame) after the solid connection.

{6 |||||| 5 ||||| 4 |||| 3 ||| 1 || 0 | -11}

We get the associated infinitesimal by first chilling, i.e., by subtracting 1 pt. for each Black stone played and adding 1 pt. for each White stone played.

{5 |||||| 5 ||||| 5 |||| 5 ||| 4 || 4 | -5}

Then we subtract the mean value. That is, the chilled value = 5 + I where I =

{0 |||||| 0 ||||| 0 |||| 0 ||| -1 || -1 | -10}

RobertJasiek · Post by **RobertJasiek** » Thu Nov 02, 2017 9:43 am

Reading https://senseis.xmp.net/?Count I thought I would have understood traversal (what confusingly that page and CGT calls reversal). However, when trying to apply it to the following examples, I notice that I have understood nothing.

Code: Select all

     B
    / \
D(1)   E
      / \
     F   G(-13)
    / \
H(2)   I(-10)

Code: Select all

          A
         / \
        B   C(-13)
       / \
   D(1)   E
         / \
        F   G(-13)
       / \
   H(2)   I(-10)

The second example is one move earlier than the first example. The second example can be represented as what looks like a hane-and-connect "sente" sequence.

How to distinguish and identify local gote, simple local sente and traversal from each other? What is the exact general procedure? Which tentative or final - gote or sente - counts and move values to calculate for which nodes? How and procedurally when? Which conditions determine the initial positions' types? What distinguishes a long sente sequence (more than 2 moves) from a traversal sequence?

I understand the conditions for simple local gote and simple local sente. However, when I try to apply them to the examples, I am confused.

Bill Spight · Post by **Bill Spight** » Thu Nov 02, 2017 10:05 am

RobertJasiek wrote:Reading https://senseis.xmp.net/?Count I thought I would have understood traversal (what confusingly that page and CGT calls reversal). However, when trying to apply it to the following examples, I notice that I have understood nothing.
Code: Select all
     B
    / \
D(1)   E
      / \
     F   G(-13)
    / \
H(2)   I(-10)
Code: Select all
          A
         / \
        B   C(-13)
       / \
   D(1)   E
         / \
        F   G(-13)
       / \
   H(2)   I(-10)
The second example is one move earlier than the first example. The second example can be represented as what looks like a hane-and-connect "sente" sequence.

How to distinguish and identify local gote, simple local sente and traversal from each other? What is the exact general procedure? Which tentative or final - gote or sente - counts and move values to calculate for which nodes? How and procedurally when? Which conditions determine the initial positions' types? What distinguishes a long sente sequence (more than 2 moves) from a traversal sequence?

I understand the conditions for simple local gote and simple local sente. However, when I try to apply them to the examples, I am confused.

Reverses sometimes catch even advanced players.

I plead guilty, myself.

B is easy. It equals J.

Code: Select all

     J
    / \
D(1)   I(-10)

The reason is that H(2) > D(1).

But in A, C(-13) = G(-13), which suggests that A = E. OTOH, maybe A = K.

Code: Select all

          K
         / \
        J   C(-13)
       / \
   D(1)   I(-10)

Both are 3 pt. sente for Black, but E has a bigger threat.

We can do difference games to find the answer, but since E carries a larger threat than K, White should not play from B to E without the intention of continuing to I. Sans ko, the play to H will not occur with correct play, but the play to D might. So A = K.

The easy way to get there is to work bottom up.

RobertJasiek · Post by **RobertJasiek** » Thu Nov 02, 2017 11:38 am

Different solution attacks for each example make me unhappy. I want the generally applicable procedure!

I have also come up with the idea of using bottom-up but have not figured out yet how to do so in general.

MGE gives the following condition: If Black plays from A (to B) and White replies to C so that A >= C, then B-C can be pruned by letting the children of C be children of A.

By symmetry, I conclude: If White plays from A (to B) and Black replies to C so that A <= C, then B-C can be pruned by letting the children of C be children of A.

However, how to determine what is the count of A? I know, I know, bottom-up. Ugh. Will try. My problem remains: what distinguishes traversal from long sente sequences? More later.

Bill Spight · Post by **Bill Spight** » Thu Nov 02, 2017 11:48 am

In this example it is not true that E <= A, despite appearances, so the play from A to B does not reverse.

The question does not depend upon the count of A; the count of A is indisputably 10, in either case.

Life In 19x19

Sente, gote and endgame plays

Re: Sente, gote and endgame plays

Re: Sente, gote and endgame plays

Re: Sente, gote and endgame plays

Re: Sente, gote and endgame plays

Re: Sente, gote and endgame plays

Re: Sente, gote and endgame plays

Re: Sente, gote and endgame plays

Re: Sente, gote and endgame plays

Re: Sente, gote and endgame plays

Re: Sente, gote and endgame plays

Re: Sente, gote and endgame plays

Re: Sente, gote and endgame plays

Re: Sente, gote and endgame plays

Re: Sente, gote and endgame plays

Re: Sente, gote and endgame plays