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 Post subject: Re: Sente, gote and endgame plays
Post #121 Posted: Sat Aug 19, 2017 10:30 am 
Honinbo

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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?

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 Post subject: Re: Sente, gote and endgame plays
Post #122 Posted: Sat Aug 19, 2017 10:38 am 
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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.

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 Post subject: Re: Sente, gote and endgame plays
Post #123 Posted: Sat Aug 19, 2017 11:11 am 
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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 :w1: 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 :w9: as sente she would get only 40 points, so she does not answer it. However, :w11: is sente, and Black answers it. So is :w15:.

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. :)

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 Post subject: Re: Sente, gote and endgame plays
Post #124 Posted: Sat Aug 19, 2017 11:25 am 
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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?!


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 Post subject: Re: Sente, gote and endgame plays
Post #125 Posted: Sat Aug 19, 2017 11:39 am 
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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. :) :clap:

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 Post subject: Re: Sente, gote and endgame plays
Post #126 Posted: Sat Aug 19, 2017 11:57 am 
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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.

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 Post subject: Re: Sente, gote and endgame plays
Post #127 Posted: Sat Aug 19, 2017 4:29 pm 
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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.

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 Post subject: Re: Sente, gote and endgame plays
Post #128 Posted: Sat Aug 19, 2017 5:10 pm 
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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?

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 Post subject: Re: Sente, gote and endgame plays
Post #129 Posted: Sat Aug 19, 2017 5:37 pm 
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Yes that does matter. You can have a look here viewtopic.php?f=15&t=14292
where Bill explained some of it to me

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 Post subject: Re: Sente, gote and endgame plays
Post #130 Posted: Sat Aug 19, 2017 10:52 pm 
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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?

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 Post subject: Re: Sente, gote and endgame plays
Post #131 Posted: Sun Aug 20, 2017 1:34 am 
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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.

Quote:
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}

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 Post subject: Re: Sente, gote and endgame plays
Post #132 Posted: Thu Nov 02, 2017 9:43 am 
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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:
     B
    / \
D(1)   E
      / \
     F   G(-13)
    / \
H(2)   I(-10)



Code:
          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.

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 Post subject: Re: Sente, gote and endgame plays
Post #133 Posted: Thu Nov 02, 2017 10:05 am 
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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:
     B
    / \
D(1)   E
      / \
     F   G(-13)
    / \
H(2)   I(-10)



Code:
          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:
     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:
          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. :)

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 Post subject: Re: Sente, gote and endgame plays
Post #134 Posted: Thu Nov 02, 2017 11:38 am 
Judan

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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.

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 Post subject: Re: Sente, gote and endgame plays
Post #135 Posted: Thu Nov 02, 2017 11:48 am 
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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.

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 Post subject: Re: Sente, gote and endgame plays
Post #136 Posted: Thu Nov 02, 2017 12:15 pm 
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Code:
     B
    / \
D(1)   E
      / \
     F   G(-13)
    / \
H(2)   I(-10)


I work myself bottom-up:

Code:
     B
    / \
D(1)   E
      / \
F(-4)   G(-13)
    / \
H(2)   I(-10)


The gote count of F is (2 + (-10)) / 2 = -4. F is a simple gote, so the gote count is its count.

Code:
     B
    / \
D(1)   E(-8.5)
      / \
F(-4)   G(-13)
    / \
H(2)   I(-10)


The tentative gote count of E is (-4 + (-13)) / 2 = -8.5.

The tentative gote move value of E is 4.5. At the child F, the follow-up (gote) move value is 6.

The tentative gote move value of E is smaller than the follow-up move value: 4.5 < 6. This condition identifies a local sente.

Therefore, we discard the tentative values of E but determine the sente count of E as the count of the sente follower: -10.

Code:
     B
    / \
D(1)   E(-10)
      / \
F(-4)   G(-13)
    / \
H(2)   I(-10)


The tentative gote count of B is (1 + (-10)) / 2 = -4.5.

Code:
     B(-4.5)
    / \
D(1)   E(-10)
      / \
F(-4)   G(-13)
    / \
H(2)   I(-10)


The tentative gote move value of B is 5.5. At the child E, the follow-up move value is what?

E is a local sente with the sente move value 3. So the follow-up move value is, I think, a sente move value and is 3.

However, why may we compare the tentative gote move value of B to the follow-up sente move value at the child E?

Should we not rather, or also(?!), compare to the black gote move value from E to F, which is 6?

Now, maybe this is a case of checking for traversal? We have B < F <=> -4.5 < -4 in a White-Black sequence so traversal applies, E-F may be pruned and we traverse to I. Hence we get the following diagram.

I have been unsure about possibly checking traversal. Why would this be the right thing to do?

Code:
     B
    / \
D(1)   I(-10)


This is what you suggested, except that you have replaced B by J. Why?

Now, this simplification tree looks like a simple gote. But is it? Or does it represent another type of local endgame that we should call a "traversal" rather than a "local gote"?

Now, what are the correct count and move value of the simplified game?

I will try the other example tree later.

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 Post subject: Re: Sente, gote and endgame plays
Post #137 Posted: Thu Nov 02, 2017 1:23 pm 
Honinbo

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


I work myself bottom-up:

Code:
     B
    / \
D(1)   E
      / \
F(-4)   G(-13)
    / \
H(2)   I(-10)


The gote count of F is (2 + (-10)) / 2 = -4. F is a simple gote, so the gote count is its count.

Code:
     B
    / \
D(1)   E(-8.5)
      / \
F(-4)   G(-13)
    / \
H(2)   I(-10)


The tentative gote count of E is (-4 + (-13)) / 2 = -8.5.

The tentative gote move value of E is 4.5. At the child F, the follow-up (gote) move value is 6.

The tentative gote move value of E is smaller than the follow-up move value: 4.5 < 6. This condition identifies a local sente.

Therefore, we discard the tentative values of E but determine the sente count of E as the count of the sente follower: -10.

Code:
     B
    / \
D(1)   E(-10)
      / \
F(-4)   G(-13)
    / \
H(2)   I(-10)


The tentative gote count of B is (1 + (-10)) / 2 = -4.5.

Code:
     B(-4.5)
    / \
D(1)   E(-10)
      / \
F(-4)   G(-13)
    / \
H(2)   I(-10)


The tentative gote move value of B is 5.5. At the child E, the follow-up move value is what?


That's not exactly the right question.

Quote:
E is a local sente with the sente move value 3. So the follow-up move value is, I think, a sente move value and is 3.


In an actual go game, the play from B to E gains 5.5 pts., but then the local temperature drops to 3, as you point out. So Black may well stop there and not continue to F. The reason is that the play from E to F may become a ko threat. The slight risk is that White may be able to play the reverse sente to G. So normally Black will stop local play and play elsewhere.

But finding the mean value is a local matter, except for ko positions.

Quote:
However, why may we compare the tentative gote move value of B to the follow-up sente move value at the child E?

Should we not rather, or also(?!), compare to the black gote move value from E to F, which is 6?


Yes, that is what we do. :)

Quote:
Now, maybe this is a case of checking for traversal? We have B < F <=> -4.5 < -4 in a White-Black sequence so traversal applies, E-F may be pruned and we traverse to I. Hence we get the following diagram.

I have been unsure about possibly checking traversal. Why would this be the right thing to do?


If you want to simplify, it is the right thing to do.

Is F >= B? That is the question.

Code:
     B(-4.5)            +           -F
    / \                             / \
D(1)   E(-10)                     10   -2
      / \
F(-4)   G(-13)
    / \
H(2)   I(-10)


If so, then B - F <= 0, which means that if Black plays first in the combination above, White can win or get jigo in an even number of plays.

First, suppose that Black plays in -F to 10, then White replies B -> E -> F -> -10, for a result of 0 in an even number of plays.

Second, suppose that Black plays in B to 1, then White replies in -F to -2, for a result of -1; White wins.

So F >= B and the play, B -> E reverses to I(-10). :)

Quote:
Code:
     B
    / \
D(1)   I(-10)


This is what you suggested, except that you have replaced B by J. Why?


Because B and J are different games. The question is whether they are equal. They are. :)

Quote:
Now, this simplification tree looks like a simple gote. But is it? Or does it represent another type of local endgame that we should call a "traversal" rather than a "local gote"?


The simplified game, J, is a simple gote. B is not. As we have seen, in a real game Black is likely to leave the sente at E on the board for a while. :)

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 Post subject: Re: Sente, gote and endgame plays
Post #138 Posted: Thu Nov 02, 2017 10:40 pm 
Judan

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


The gote count of F is (2 + (-10)) / 2 = -4. F is a simple gote, so the gote count is its count.

Code:
          A
         / \
        B   C(-13)
       / \
   D(1)   E(-8.5)
         / \
    F(-4)   G(-13)
       / \
   H(2)   I(-10)


The tentative gote count of E is (-4 + (-13)) / 2 = -8.5.

The tentative gote move value of E is 4.5. At the child F, the follow-up gote move value is 6.

The tentative gote move value of E is smaller than the follow-up move value: 4.5 < 6. This condition identifies a local sente.

Therefore, we discard the tentative values of E but determine the sente count of E as the count of the sente follower: -10.

Code:
          A
         / \
        B   C(-13)
       / \
   D(1)   E(-10)
         / \
    F(-4)   G(-13)
       / \
   H(2)   I(-10)


The tentative gote count of B is (1 + (-10)) / 2 = -4.5.

Code:
          A
         / \
  B(-4.5)   C(-13)
       / \
   D(1)   E(-10)
         / \
    F(-4)   G(-13)
       / \
   H(2)   I(-10)


The tentative gote move value of B is 5.5.

E is a local sente with the follow-up sente move value 3.

The tentative gote move value of B is larger than the follow-up sente move value at the child E, that is, 5.5 > 3. This condition identifies a local gote. (Besides, the follow-up gote move value at the child D is 0 for a pass. The tentative gote move value of B is larger than the follow-up gote move value at the child D, that is, 5.5 > 0. This condition also identifies a local gote.)

So B is a local gote and we keep its tentative gote move value as its gote move value.

We can check for traversal: We have B <= F <=> -4.5 <= -4 in a White-Black sequence so traversal applies, E-F may be pruned and we traverse to I. Hence we get the following equal diagram.

Code:
          A'
         / \
  B(-4.5)   C(-13)
       / \
   D(1)   I(-10)


The line from B to I represents the sequence (move to E) - (move to F) - (move to I).

In the simplified game, the tentative gote count of A' is -8,75.

Code:
          A'(-8.75)
         / \
  B(-4.5)   C(-13)
       / \
   D(1)   I(-10)


In the simplified game, the tentative gote move value of A' is 4.25. At the child B, the follow-up gote move value is 5.5.

The tentative gote move value of A' is smaller than the follow-up gote move value, that is 4.25 < 5.5. This condition identifies a local sente.

Therefore, we discard the tentative values of A' but determine the sente count of A' as the count of the sente follower I: -10.

Code:
          A'(-10)
         / \
  B(-4.5)   C(-13)
       / \
   D(1)   I(-10)


The local endgame A' is a local sente with the already calculated sente count -10 and the sente move value 3.

Since A = A', we derive the values for original game A:

Code:
          A(-10)
         / \
  B(-4.5)   C(-13)
       / \
   D(1)   E(-10)
         / \
    F(-4)   G(-13)
       / \
   H(2)   I(-10)


The local endgame A has the sente count -10 and the sente move value 3.

***

Let us go back to the analysis of A before we used traversal to simplify the game:

Code:
          A
         / \
  B(-4.5)   C(-13)
       / \
   D(1)   E(-10)
         / \
    F(-4)   G(-13)
       / \
   H(2)   I(-10)


We have already identified B as a local gote with gote count -4.5 and the gote move value 5.5.

Now, we proceed to analyse A.

The local endgame A has the tentative gote count -8.75.

Code:
          A(-8.75)
         / \
  B(-4.5)   C(-13)
       / \
   D(1)   E(-10)
         / \
    F(-4)   G(-13)
       / \
   H(2)   I(-10)


The tentative gote move value of A is 4.25. At the child B, the follow-up gote move value is 5.5.

The tentative gote move value of A is smaller than the follow-up gote move value, that is 4.25 < 5.5. This condition identifies a local sente.

Therefore, we discard the tentative values of A but determine the sente count of A as the count of the sente follower E: -10.

Code:
          A(-10)
         / \
  B(-4.5)   C(-13)
       / \
   D(1)   E(-10)
         / \
    F(-4)   G(-13)
       / \
   H(2)   I(-10)


The local endgame A is a local sente with the sente count -10 and the sente move value 3.

Locally, without ko threat play, the sequence from A to I is traversed as Black's sente. The sente move value 3 at A and E and the follow-up move values 5.5 at B and 6 at F must be considered for the timing in the context of the global temperature.

***

For the local endgame A, traversal provides an optional technique with which type, count and move value of A can be determined. The type is a local sente. Traversal describes that, once Black has started the local sente sequence at a suitable global temperature, it may as well traverse to I.

Since the count of C equals the count of G, Black stopping at A incurs the same risk as stopping at E that White plays reverse sente. Since the follow-up move value of B is 5.5 and smaller than the follow-up move value 6 at F, if White should not stop at B allowing Black to achieve the count 1 at D, White should even less stop at F allowing Black to achieve the larger count 2 at H.

Ta-ta--ta---taaa!

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 Post subject: Re: Sente, gote and endgame plays
Post #139 Posted: Thu Nov 02, 2017 11:14 pm 
Judan

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(Continuing discussion from my previous post.)

Bill Spight wrote:
RobertJasiek wrote:
Code:
     B
    / \
D(1)   E
      / \
     F   G(-13)
    / \
H(2)   I(-10)



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



B is easy. It equals J.

Code:
     J
    / \
D(1)   I(-10)


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


You sound as if this was sufficient reason for B equalling J. Please explain. Is such a condition general for all trees? If so, what is the general condition, that is, what counts to compare how to assess as sufficient reason that a local endgame simplified by traversal equals its simplified game?

Quote:
We can do difference games to find the answer


To check that a local endgame and its traversal-simplification are equal? Why? This is a truth we (eh, should) know from CGT.

Quote:
the count of A is indisputably 10


I'd even say it is indisputably -10. ;)

Quote:
Quote:
However, why may we compare the tentative gote move value of B to the follow-up sente move value at the child E?
Should we not rather, or also(?!), compare to the black gote move value from E to F, which is 6?

Yes, that is what we do.


I still wonder for what purposes exactly we a) compare the tentative gote move value of B to the follow-up sente move value at the child E and b) compare the tentative gote move value of B to the black gote move value from E to F.

Quote:
Quote:
Code:
     B(-4.5)
    / \
D(1)   E(-10)
      / \
F(-4)   G(-13)
    / \
H(2)   I(-10)



B is not [a simple gote.]


Let me see: B has the tentative gote move value 5.5 and the tentative sente move value D(1) - F(-4) = 5. The tentative gote move value 5.5 is larger than the tentative sente move value 5. This condition identifies a local sente. So B is White's local sente.

Furthermore, E is Black's local sente.

Oh!

The intermediate position E is involved in both White's local sente at B and Black's local sente at E. That's an alternative explanation why we may traverse the sequence from B to I if played at an appropriate temperature! The whole sequence is White's gote sequence but its parts are either player's sente sequences. Ugh. No suprise I found things to be so complicated. Have I understood it correctly now?

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 Post subject: Re: Sente, gote and endgame plays
Post #140 Posted: Fri Nov 03, 2017 12:59 am 
Honinbo

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RobertJasiek wrote:
(Continuing discussion from my previous post.)

Bill Spight wrote:
RobertJasiek wrote:
Code:
     B
    / \
D(1)   E
      / \
     F   G(-13)
    / \
H(2)   I(-10)



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



B is easy. It equals J.

Code:
     J
    / \
D(1)   I(-10)


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


You sound as if this was sufficient reason for B equalling J. Please explain. Is such a condition general for all trees? If so, what is the general condition, that is, what counts to compare how to assess as sufficient reason that a local endgame simplified by traversal equals its simplified game?


Given: Game B
Code:
        B
       / \
      D   E
         / \
        F   G
       / \
      H   x


where B, D, E, F, G, and H are games; B, E, and F are not numbers; x is a number; and H >= D.

To prove: B -> E -> F reverses to x. I.e., F >= B <=> B - F <= 0.

Code:
        B         +        -F        <=  0
       / \                 / \
      D   E              -x  -H
         / \
        F   G
       / \
      H   x

Lines of play where Black plays first.
1) Black moves from B to D, then White moves from -F to -H. The result is D - H <= 0.
2) Black moves from -F to -x, then White moves from B to E, Black moves from E to F, and White moves from F to x. The result is x - x = 0.

When Black plays first White can always play to a final position that is less than or equal to 0.

QED.

Note that because x is a number, Black has no good move in 2) after White moves to E, except to move to F, by the number avoidance theorem.

Quote:
Quote:
We can do difference games to find the answer


To check that a local endgame and its traversal-simplification are equal? Why? This is a truth we (eh, should) know from CGT.


IMX, I have often found it easier to check that a game and its simplification are equal than to check that the conditions for reversal or dominance are met. :)

Quote:
Quote:
the count of A is indisputably 10


I'd even say it is indisputably -10. ;)


Sans doute. ;)

Quote:
Quote:
Quote:
However, why may we compare the tentative gote move value of B to the follow-up sente move value at the child E?
Should we not rather, or also(?!), compare to the black gote move value from E to F, which is 6?

Yes, that is what we do.


I still wonder for what purposes exactly we a) compare the tentative gote move value of B to the follow-up sente move value at the child E and b) compare the tentative gote move value of B to the black gote move value from E to F.


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


F is not a number, so the fact that C = G is not enough to tell us that the play reverses through E. However, the play in B reverses through F.

Quote:
No suprise I found things to be so complicated. Have I understood it correctly now?


Reverses are tricky. I think you have a good understanding. :D

_________________
The Adkins Principle:
At some point, doesn't thinking have to go on?
— Winona Adkins

Visualize whirled peas.

Everything with love. Stay safe.


Last edited by Bill Spight on Fri Nov 03, 2017 1:14 am, edited 2 times in total.
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