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 Post subject: Re: Sente, gote and endgame plays
Post #21 Posted: Thu Oct 13, 2016 8:35 am 
Honinbo

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RobertJasiek wrote:
Bill conveys a sketch of an alternative proof using the method of multiples. It seems that a full proof would rely on correct play for the limit of N->oo copies of P. By dividing the count of the ensemble by N, we get the expected average count in P. Iterating N by 1, 2 or a different number on all integers, only even or only odd integers does not affect the construction of such a proof; for a given P, we would choose the most fitting iteration step and subset of all integers. Nevertheless, I remain to be convinced why any proof using the method of multiples is well-defined.


The method of multiples has been used for many years, maybe more than a century, to find the mean value of games. IIUC it was used to define the mean value of games. (Independent go positions are games.) Here I used it to indicate that Black will choose the sente option under certain conditions, the gote option under others. That is enough to tell us how to characterize the original position. But without a formal definition of sente and gote, it is not a formal proof. (I have developed formal definitions, which is why I came up with the classification of ambiguous.) Also, the full derivation of the mean value by the method of multiples would mean showing that the mean value converges as the number of multiples goes to infinity and showing that it is the same when White plays first.

Quote:
The method of multiples is an overkill for distinguishing local gote from local sente.


Thermography is a simpler technique, but I think that the method of multiples is easier to comprehend.

Quote:
When we will have identified and calculated all possible types including tentative sente options, let me ask again: For arbitrary such positions P without ko, how to spell out a general method of distinguishing local gote, local sente and the ambiguous case?


The method of multiples works and is perfectly general. As I indicated, you can also use thermography, but you have to use colored masts to distinguish ambiguous positions. (I came up with the idea of colored masts circa 2000.)

Quote:
How much in this thread is current research and what did already exist, possibly in more general contexts, among CGT researchers or by you, Bill?


Except for not having the concept of ambiguous positions, top go players 200 years ago had ways of distinguishing between sente and gote in this type of play. :) Thermography was invented some 40 years ago. Around that time I had my own methods to reach the same conclusions (except for not yet having the ambiguous classification).

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 Post subject: Re: Sente, gote and endgame plays
Post #22 Posted: Thu Oct 13, 2016 9:05 am 
Judan

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Bill Spight wrote:
Here I used it to indicate that Black will choose the sente option under certain conditions, the gote option under others. That is enough to tell us how to characterize the original position.


Is it? We characterise a local (non-ambiguous) position as either local gote or local sente so that then we can calculate the correct count and move value. I am sure that you would also do the latter but apparently you use a different procedural step when and how to calculate them, do you?

Quote:
Thermography is a simpler technique, but I think that the method of multiples is easier to comprehend.


I think that using neither but directly distinguishing local gote from local sente is the simplest.

Quote:
top go players 200 years ago had ways of distinguishing between sente and gote in this type of play. :)


Any explicit ways?

Quote:
Thermography was invented some 40 years ago.


Thermography is a methodology with a much broader scope of application (not that I understand much of it). I have wondered about specific methods for distinguishing local gote from local sente, before we even consider environments.

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 Post subject: Re: Sente, gote and endgame plays
Post #23 Posted: Thu Oct 13, 2016 1:20 pm 
Honinbo

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RobertJasiek wrote:
Bill Spight wrote:
Here I used it to indicate that Black will choose the sente option under certain conditions, the gote option under others. That is enough to tell us how to characterize the original position.


Is it? We characterise a local (non-ambiguous) position as either local gote or local sente so that then we can calculate the correct count and move value. I am sure that you would also do the latter but apparently you use a different procedural step when and how to calculate them, do you?


When I was first learning thermography I figured out the correct plays and then drew the thermograph. Berlekamp said that that was backwards. Thermographs tell us what typically correct plays are. Go players often start by characterizing plays as sente and gote and then figuring out the count and move value. If you start by figuring out the count and move value, which is what the method of multiples does, you can avoid some mistakes and confusion. :)

Quote:
Quote:
Thermography is a simpler technique, but I think that the method of multiples is easier to comprehend.


I think that using neither but directly distinguishing local gote from local sente is the simplest.


But somewhat error prone.

Quote:
Quote:
top go players 200 years ago had ways of distinguishing between sente and gote in this type of play. :)


Any explicit ways?


Seat of the pants, as far as I can tell. ;) But usually right. :)

Quote:
Quote:
Thermography was invented some 40 years ago.


Thermography is a methodology with a much broader scope of application (not that I understand much of it). I have wondered about specific methods for distinguishing local gote from local sente, before we even consider environments.


The method of multiples does that, unless you think of the copies as an environment.

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Last edited by Bill Spight on Thu Oct 13, 2016 1:25 pm, edited 1 time in total.
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 Post subject: Re: Sente, gote and endgame plays
Post #24 Posted: Thu Oct 13, 2016 1:21 pm 
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P versus other plays

First let us simplify things a bit. Let Black have a move to a position worth G, a positive number, with gote, but let White have a move to a position worth -R, a negative number, and let Black also have a sente move to a position worth 0. Now we know that if G > R, the position is gote, if R > G, it is sente, and if G = R, it is ambiguous.

Next, let us add a play, a simple gote where Black can move to a position worth T and White can move to a position worth -T. (All variables are greater than 0.) T is smaller than BIG, so Black still has a sente play.

Black to play.

1) Black plays sente, then plays to T. Result: T.

2) Black plays to G, then White plays to -T. Result: G - T.

3) Black plays to T, then White plays to -R. Result: T - R.

T > T - R, so option 3) is out. Comparing 1) to 2) we find that if G > 2T then Black should play to G (gote) and if 2T > G Black should play sente first.

(This was my approach in the 1970s. :))

Now let us add another simple gote, where Black can move to T1 and White can move to -T1, with T >= T1 > 0.

1) Black plays sente, then plays to T, then White plays to -T1. Result: T - T1.

2) Black plays to G, then White plays to -T and Black plays to T1. Result: G - T + T1.

3) Black plays to T, then White plays to -R, then Black plays to T1. Result: -R + T + T1.

4) Black plays to T, then White plays to -T1, then Black plays sente. Result: T - T1, the same result as 1).

5) Black plays to T, then White plays to -T1, then Black plays to G. Result: G + T - T1, which is at least as good as the result for 2).

4) and 5) show that if a White reply to -T1 is correct, a Black play to T dominates the other plays. However, if 3) is inferior to 1) or 2), then Black should not play to T. So we can ignore White’s response to -T1.

Comparing 2) to 3), we find that 2) is better if G + R > 2T. I. e., if playing P as a gote is better that playing T. T1 does not enter the picture.

Comparing 1) to 2), we find that playing to G is better than the sente if G > 2(T - T1).

Comparing 1) to 3), we find that playing sente is better than playing to T if R > 2T1. T does not enter the picture.

Now let us add other simple gote similar to T and T1 to make an environment, such that T >= T1>= T2 >= . . . > 0

We still have three comparisons.

1) Black plays sente. Result: T - T1 + T2 - . . . .

2) Black plays to G. Result: G - T + T1 - T2 + . . . .

3) Black plays to T, then White plays to -R. Result: -R + T + T1 - T2 - . . . .

Comparing 2) to 3) we get that Black should not play to T if G + R > 2T. Same as before. :)

If 2T > G + R we compare 1) to 3) and get that Black should play sente if R > 2(T1 - T2 + . . .).

If G + R > 2T we compare 1) to 2) and get that Black should play gote if G > 2T - 2(T1 - T2 + . . .). I have separated T from T1, T2, etc., because T does not appear in the comparison between 1) and 3). It really should not be considered part of the environment.

We may estimate T1 - T2 + . . . as T1/2. The estimate does not affect the comparison between 2) and 3), but it does affect the others. Using the estimate we get

1) vs 3): Compare R to T1.

1) vs 2): Compare G to 2T - T1.

——

We may also consider a generalized environment with a temperature of T. In that case thermography will tell us that the comparison between 1) and 2) is between G and T. It applies when P is classified as sente or ambiguous.

My method is more powerful than thermography, as it suggests that the environment should start with T1, and it applies to the case with a losing sente, i. e., when G > R and P is classified as gote. :) The sente option does not show up in the thermograph in that case.

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 Post subject: Re: Sente, gote and endgame plays
Post #25 Posted: Thu Oct 13, 2016 4:51 pm 
Judan

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Bill Spight wrote:
Quote:
directly distinguishing local gote from local sente is the simplest.

But somewhat error prone.


What kinds of errors are you speaking of?

***

(Currently I am studying the distinction of local gote / sente. E.g., I want to search some examples. I will study value calculations also involving environments a bit later and need to catch up some very old posts and articles, too.)

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 Post subject: Re: Sente, gote and endgame plays
Post #26 Posted: Thu Oct 13, 2016 5:05 pm 
Honinbo

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RobertJasiek wrote:
Bill Spight wrote:
Quote:
directly distinguishing local gote from local sente is the simplest.

But somewhat error prone.


What kinds of errors are you speaking of?



IMX, one of the most common errors is continuing a line of play past the point where the local temperature has dropped. There are also errors of calculation, but they are typically small, so little harm is done. There are errors of misclassification, especially calling a play a (local) double sente. (Locality is implied, because the whole board is not shown.) See, for instance, http://senseis.xmp.net/?YoseErrorsInMagicOfGo .

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 Post subject: Re: Sente, gote and endgame plays
Post #27 Posted: Fri Oct 14, 2016 1:59 am 
Judan

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Code:
                   P
               /  / \
              G  /   R
                Q
               / \
             BIG  S


Example for this type of positions:

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


Click Here To Show Diagram Code
[go]$$B Black's gote option, G = -33
$$----------------------------
$$|X X X X X X . . 2 3 X . X .
$$|X X X X X O X . O 1 O X X .
$$|. . . . . O . O O X O X . .
$$|O O O O O O O O . X O X . .
$$|. . . . . O X X X X O X . .
$$|. . . . . O O O O O O X . .[/go]


Click Here To Show Diagram Code
[go]$$B Black's sente option, S = -44
$$----------------------------
$$|X X X X X X . 4 1 3 X 5 X .
$$|X X X X X O X 6 O 2 O X X .
$$|. . . . . O . O O X O X . .
$$|O O O O O O O O . X O X . .
$$|. . . . . O X X X X O X . .
$$|. . . . . O O O O O O X . .[/go]


Click Here To Show Diagram Code
[go]$$W White starts, R = -48 1/3
$$----------------------------
$$|X X X X X X . . . 1 X . X .
$$|X X X X X O X . O . O X X .
$$|. . . . . O . O O X O X . .
$$|O O O O O O O O . X O X . .
$$|. . . . . O X X X X O X . .
$$|. . . . . O O O O O O X . .[/go]


Conditions:

The local endgame is a local gote if (G + R)/2 > S.
The local endgame is a local sente if (G + R)/2 < S.

Application:

(G + R)/2 > S <=> (-33 + (-48 1/3)) / 2 > -44 <=> -81 1/3 / 2 > -44 <=> -40 2/3 > -44.
The local endgame is local gote. Usually, Black chooses the gote sequence.


This post by RobertJasiek was liked by: Bill Spight
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 Post subject: Re: Sente, gote and endgame plays
Post #28 Posted: Fri Oct 14, 2016 7:19 am 
Honinbo

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Thanks. A very nice example. :)

Here is a close call.

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

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 Post subject: Re: Sente, gote and endgame plays
Post #29 Posted: Sun Nov 13, 2016 8:01 am 
Judan

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


Since nobody has done the exercise yet, here is my call. I use the obvious locale for the counts.

Click Here To Show Diagram Code
[go]$$B Black's gote option, G = (GB + GW) / 2 = -33.5
$$----------------------------
$$|X X X X X X . . 2 3 X . X .
$$|X X X X X O X . O 1 O X X .
$$|. . . . . O . O O X O X . .
$$|O O O O O O O O . X O X . .
$$|. . . . . . . O O . O X . .
$$|. . . . . . . . O . O X . .
$$|. . . . . . . . O O O X . .[/go]


Click Here To Show Diagram Code
[go]$$B G's black follower, GB = -33
$$----------------------------
$$|X X X X X X . . O X X . X .
$$|X X X X X O X . O X O X X .
$$|. . . . . O . O O X O X . .
$$|O O O O O O O O . X O X . .
$$|. . . . . . . O O 1 O X . .
$$|. . . . . . . . O . O X . .
$$|. . . . . . . . O O O X . .[/go]


Click Here To Show Diagram Code
[go]$$W G's white follower, GW = -34
$$----------------------------
$$|X X X X X X . . O X X . X .
$$|X X X X X O X . O X O X X .
$$|. . . . . O . O O X O X . .
$$|O O O O O O O O . X O X . .
$$|. . . . . . . O O 1 O X . .
$$|. . . . . . . . O . O X . .
$$|. . . . . . . . O O O X . .[/go]


Click Here To Show Diagram Code
[go]$$B Black's sente option, S = -38
$$----------------------------
$$|X X X X X X . 4 1 3 X 5 X .
$$|X X X X X O X 6 O 2 O X X .
$$|. . . . . O . O O X O X . .
$$|O O O O O O O O . X O X . .
$$|. . . . . . . O O . O X . .
$$|. . . . . . . . O . O X . .
$$|. . . . . . . . O O O X . .[/go]


Click Here To Show Diagram Code
[go]$$W White starts, R = -42 1/3
$$----------------------------
$$|X X X X X X . . . 1 X . X .
$$|X X X X X O X . O . O X X .
$$|. . . . . O . O O X O X . .
$$|O O O O O O O O . X O X . .
$$|. . . . . . . O O . O X . .
$$|. . . . . . . . O . O X . .
$$|. . . . . . . . O O O X . .[/go]


Click Here To Show Diagram Code
[go]$$B intermediate position Q = (QB + QW) / 2 = -45 / 2 = -22.5
$$----------------------------
$$|X X X X X X . . 1 3 X . X .
$$|X X X X X O X . O 2 O X X .
$$|. . . . . O . O O X O X . .
$$|O O O O O O O O . X O X . .
$$|. . . . . . . O O . O X . .
$$|. . . . . . . . O . O X . .
$$|. . . . . . . . O O O X . .[/go]


Click Here To Show Diagram Code
[go]$$B Q's black follower, QB = -7, move 1 is move 4 as continuation
$$----------------------------
$$|X X X X X X . 1 X X X . X .
$$|X X X X X O X . O O O X X .
$$|. . . . . O . O O X O X . .
$$|O O O O O O O O . X O X . .
$$|. . . . . . . O O . O X . .
$$|. . . . . . . . O . O X . .
$$|. . . . . . . . O O O X . .[/go]


Click Here To Show Diagram Code
[go]$$W Q's white follower, QW = S = -38, move 1 is move 4 as continuation
$$----------------------------
$$|X X X X X X . 1 X X X 2 X .
$$|X X X X X O X 3 O O O X X .
$$|. . . . . O . O O X O X . .
$$|O O O O O O O O . X O X . .
$$|. . . . . . . O O . O X . .
$$|. . . . . . . . O . O X . .
$$|. . . . . . . . O O O X . .[/go]


We need to do three things:
1) verify that the supposed gote sequence to G is a gote sequence indeed,
2) verify that the supposed sente sequence to S is a sente sequence indeed,
3) identify whether the local endgame is a local gote, ambiguous or local sente.

(1) The supposed gote sequence to G is a gote sequence because the move value of the follow-up is 0.5 and therefore much smaller than the gote move value of the initial position, which is (G - R) / 2 = (-33.5 - (-42 1/3)) / 2 = (-33.5 + 42 1/3)) / 2 = (8 5/6) / 2 = 4 5/12.

(2) The supposed sente sequence to S is a sente sequence due to the sente condition S < (Q + R) / 2
<=> -38 < (-22.5 + (-42 1/3)) / 2 <=> -38 < (-64 5/6) / 2 <=> -38 < -32 5/12.

(3) The local endgame is a local gote due to the condition S < (G + R) / 2
<=> -38 < (-33.5 + (-42 1/3)) / 2 <=> -38 < (-75 5/6) / 2 <=> -38 < -37 11/12.

***

Bill Spight wrote:
Suppose that the ambient temperature is T, i. e., that the gain from making the largest play elsewhere on the board is T.
[...] Then we can estimate the gain from playing out the rest of the board as T/2.


Why, and by which proof, T/2?

I am reminded of arguments related to komi compensating the right of moving first by being half the miai value of the first move. That was ca. 15 years ago, so I do not recall the proof by heart. But my real concern here is that you presume move values T >= T1 >= T2.. >= Tn > 0 and I wonder how, and due to which assumption of move value decrements, to sum up to get the excess value T/2.

Quote:
Exception 2: In a gote position to play the sente option.

This exception requires a large gote (ending in a local count of A if you play, B if your opponent plays, A > B).


Here, I do not understand what gote you mean. Is this
- the gote option G of the local endgame P, with A being the count of G's black follower and B being the count of G's white follower,
- a big gote in the environment, where there are also other, significantly smaller gotes with values T and smaller,
- something else?

Quote:
Case 1. The sente threat is at least as large as the other gote: H - S >= A - B.

Conditions:

1) S - R > T

2) A - B > G - S + T

Case 2. The sente threat is smaller than the other gote: A - B > H - S

Conditions:

1) H - G > T

2) H - R > A - B + T



Why does the term A - B occur? The counts A and B are of a (large) gote, so what is A - B? The gote's count is (A + B) / 2 and move value is (A - B) / 2. Have you already cancelled division by 2? If so, from which unequations do you start?

Why do we need A and B at all? Is it not sufficient to use one parameter for either the (large) gote's count or move value, without referring to its followers' counts A and B?


Quote:
let White have a move to a position worth -R, a negative number


Is -R negative or is R negative and you'd better write R < 0? The annotation -R confuses me here.

***

EDIT 4: overlooked the tiny follow-up. Now corrected. Oops. Fractions. Once more corrected. Then forgot to alter the gote sequence verication.

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 Post subject: Re: Sente, gote and endgame plays
Post #30 Posted: Sun Nov 13, 2016 9:57 am 
Judan

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Now, after 4 corrections, I might have solved Bill's exercise, provided the theory is right. :)

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 Post subject: Re: Sente, gote and endgame plays
Post #31 Posted: Sun Nov 13, 2016 7:12 pm 
Honinbo

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


Since nobody has done the exercise yet, here is my call. I use the obvious locale for the counts.

Click Here To Show Diagram Code
[go]$$B Black's gote option, G = (GB + GW) / 2 = -33.5
$$----------------------------
$$|X X X X X X . . 2 3 X . X .
$$|X X X X X O X . O 1 O X X .
$$|. . . . . O . O O X O X . .
$$|O O O O O O O O . X O X . .
$$|. . . . . . . O O . O X . .
$$|. . . . . . . . O . O X . .
$$|. . . . . . . . O O O X . .[/go]


This is orthodox play. :)

RobertJasiek wrote:
Bill Spight wrote:
Exception 2: In a gote position to play the sente option.

This exception requires a large gote (ending in a local count of A if you play, B if your opponent plays, A > B).


Here, I do not understand what gote you mean. Is this
- the gote option G of the local endgame P, with A being the count of G's black follower and B being the count of G's white follower,
- a big gote in the environment, where there are also other, significantly smaller gotes with values T and smaller,
- something else?


I meant another gote.

BTW, for this type of discussion, all we know about plays in the environment is that their miai values are less than the ambient temperature of the environment. The large gote is part of the foreground, not part of the background (environment).

RobertJasiek wrote:
Bill Spight wrote:
Suppose that the ambient temperature is T, i. e., that the gain from making the largest play elsewhere on the board is T.
[...] Then we can estimate the gain from playing out the rest of the board as T/2.


Why, and by which proof, T/2?

I am reminded of arguments related to komi compensating the right of moving first by being half the miai value of the first move. That was ca. 15 years ago, so I do not recall the proof by heart. But my real concern here is that you presume move values T >= T1 >= T2.. >= Tn > 0 and I wonder how, and due to which assumption of move value decrements, to sum up to get the excess value T/2.


The best estimate is an empirical question. However, the main thing that can throw the T/2 estimate off is a relatively large drop in temperature at some point, larger than the average drop in temperature, after adjusting for miai. For example, my last play problems are based upon the drop in temperature of 1 point, from temperature 1 to temperature 0. Often I throw in some miai positions of lower temperature, but I think that people have caught on to that trick by now. ;) If there is one gote at temperature 1 and a miai at temperature 3/4 and another miai at temperature 1/2, the average temperature drop is 1/3, but it is 1 after adjusting for the miai. :)

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 Post subject: Re: Sente, gote and endgame plays
Post #32 Posted: Mon Nov 14, 2016 12:02 am 
Judan

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In a peaceful environment, the values of the non-zero temperature drops are a constant D and let the first versus second in-the-environment moving players gain the excess D as close to an equal number of times as possible. If there are an even number of excess D drops, the first moving player gains the total excess 0. If there are an odd number of excess D drops, the first moving player gains the total excess D. On empirical average for such an environment, the first moving player gains the total excess D/2, independent of the initial ambient temperature T.

Is such an environment realistic? Why should be an environment with the total excess T/2 for the first in-the-environment moving player be much more realistic empirically?

Suppose D is small and the environment consists of two phases: phase I has almost constant temperature T but with, on empirical average, the first moving player gaining the total excess D/2 - phase II starts after a significant temperature drop to temperature U, phases out the game ending at the smallest temperature O and lets, on empirical average, the continuing moving player gain the total excess D/2. Since D is small, we can simplify by ignoring D/2 for either phase. Phase I can have an even or odd number of moves. If the number is even, the first moving player gains the simplified excess 0 during phase I. If the number is odd, the first moving player gains the simplified excess T during phase I. On empirical average, for an unknown parity of the number of moves of phase I, the first moving player gains the simplified excess T/2. Suppose, on empirical average, U = T/2 so the drop to phase II is T/2. If phase I has an even number of moves, the first moving player gains the simplified excess 0 during phase I and excess T/2 by gaining the drop to phase II. If phase I has an odd number of moves, the first moving player gains the simplified excess T during phase I and the second moving player gains T/2 (i.e., lets the first moving player lose T/2) from the drop to phase II; the first moving player's excess is T - T/2 = T/2. Hence, regardless of the parity of the number of moves in phase I, in total, the first moving player gains the simplified excess T/2.

This presumes that there are only two phases. How about several phases?

How does the calculation change if the phase drop differs from T/2?

What changes if the drops within phase I are significant instead of a small, essentially constant D? (For phase II, such a D is a reasonable assumption in many practical cases during the early endgame.)

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Post #33 Posted: Mon Nov 14, 2016 7:24 am 
Judan

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Here is the basic form of an ideal environment:

Suppose an environment with temperatures from T to 0, apart from (possibly multiple) miai pairs, dropping constantly at N>>0 drops and the players profitting from drops equally often or, in the case of an odd number of drops, the first moving player gaining once more in alternating play.

Each drop is T/N.

For N even, each pair of moves with values V and V-T/N, the first moving player gains T/N. For all N/2 pairs, the first moving player gains (N/2)*(T/N) = T/2.

For N odd, the first moving player gets T/N once and then there are (N-1)/2 pairs:
T/N + ((N-1)/2)*(T/N) = T/N + (N-1)*T/2N = T/N + N*T/2N - T/2N = T/2 + T/2N

On empirical average of the parity cases, the first moving player gains T/2 + T/4N ~= T/2 as a good approximation.

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 Post subject: Re: Sente, gote and endgame plays
Post #34 Posted: Mon Nov 14, 2016 9:52 am 
Honinbo

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The go literature talks about the importance of getting the last play at three points in the game, the last big play of the opening, the last large endgame play, and the last play of the game. The third one has been studied by mathematicians. It is the easiest to demonstrate. IMO, there is something to the first one. Going by komi of around 7 points, we can surmise that the first play gains around 14 points. But it could well be around 16 points, and the reason that komi is not higher is that the symmetry of the empty go board yields miai for the first several moves. There could be a temperature drop of as much as a point or two after the last big play of the opening. This is speculation, because it is not easy to determine how much an opening play gains. As for the last large yose, my sense is that sometimes there is a significant temperature drop afterwards, sometimes not. Certainly I have won a number of games at the large endgame stage, which involved getting the last large yose. :) Even though it is in general difficult to calculate large endgame plays with accuracy, you can get good approximations. It might be interesting to study getting the last large yose in pro games. :)

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 Post subject: Re: Sente, gote and endgame plays
Post #35 Posted: Sat Dec 10, 2016 12:53 am 
Judan

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Quotation reference: viewtopic.php?p=200624#p200624

John Fairbairn wrote:
O Meien [...] a formula for predicting certainty of victory. [...] obtain the "value of a move" by "absolute counting"

Advantage of first move = half the value of a move
Margin of error = half the advantage of first move

[...]
* If it is the opponent’s turn to play, even if you add the advantage of first move and the margin of error to the opponent’s territory, if you are ahead you have “certainty.”

* If there is an outstanding big move for the opponent, assume he can play there then count. Add the advantage of first move to you territory and add the margin of error to the opponent’s territory.


Without background explanation of theory, O's formula might appear to lack justification. An outstanding big move is just a special case; we may as well assume an ensemble of big moves to be dissolved before then applying the formula to the remaining peaceful environment with T being the ambient temperature, that is the miai value of the largest move in the environment. O suggests that the advantage of this first move is half the value of a move, that is, T/2. For an ideal environment, I have proven that this is a good approximation in viewtopic.php?p=213050#p213050 and more precisely, on empirical average of the parity cases, the first moving player gains T/2 + T/4N for N>>0 available drops in the environment.

Then O's margin of error, which he specifies as half the advantage of first move, is T/4. However, T/4 >> T/4N for N>>0. For the sake of choosing moves, his margin of error is unnecessarily large. It serves a different purpose: defensive positional judgement. By adding T/4 to the opponent's count, a player, according to O, could know with "certainty" to be leading. As an empirical statement, it makes sense now. OTOH, unlike the advantage of starting in the environment ("having the first move"), the margin of error is an arbitrary amount as long as N is large. We might as well not use a specific margin of error but be aware that winning certainty increases with the size of a player's whole board count.

O's margin of error T/4 may be pragmatic but it suggests a black/white picture on having / not having the certainty of winning. This is unnecessarily simplistic because we speak of estimates of the whole board count - not of knowing for sure whether Black or White wins if an estimate trespasses the rather arbitrary T/4 value line. We must also recall that T/2 (and thus T/4) is derived for ideal envionments. In practice, environments need not be ideal, so we cannot set a precise boundary, such as T/4, to distinguish certainty from uncertainty of winning. T/4 must be understood as a guideline. Like every margin of error, we should write ~= T/4 and say "more certain" for larger counts versus "less certain" for smaller counts - instead of declaring "certainty".

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 Post subject: Re: Sente, gote and endgame plays
Post #36 Posted: Thu Dec 29, 2016 12:33 pm 
Judan

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Bill Spight wrote:
Now let us add other simple gote similar to T and T1 to make an environment, such that T >= T1>= T2 >= . . . > 0

We still have three comparisons.

1) Black plays sente. Result: T - T1 + T2 - . . . .

2) Black plays to G. Result: G - T + T1 - T2 + . . . .

3) Black plays to T, then White plays to -R. Result: -R + T + T1 - T2 - . . . .

Comparing 2) to 3) we get that Black should not play to T if G + R > 2T. Same as before. :)

If 2T > G + R we compare 1) to 3) and get that Black should play sente if R > 2(T1 - T2 + . . .).

If G + R > 2T we compare 1) to 2) and get that Black should play gote if G > 2T - 2(T1 - T2 + . . .). I have separated T from T1, T2, etc., because T does not appear in the comparison between 1) and 3). It really should not be considered part of the environment.

We may estimate T1 - T2 + . . . as T1/2. The estimate does not affect the comparison between 2) and 3), but it does affect the others. Using the estimate we get

1) vs 3): Compare R to T1.

1) vs 2): Compare G to 2T - T1.


IIYC, your first steps of comparison clarify whether Black should start by taking T and you rely on T ~= T1 to suggest approximating conditions for the comparisons (1) to (3) and (2) to (3).

However, you completely lose me with your comparison (1) to (2); why do you compare G to 2T - T1? (1) reduces to T/2 and (2) is G - T/2. Comparing these two terms gives G >?< T.

Another aspect I have not completely understood yet is why we need not consider the strategy of Black taking T then White taking T1 for the sake of deciding whether Black should start by taking T.

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Post #37 Posted: Thu Dec 29, 2016 2:26 pm 
Honinbo

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RobertJasiek wrote:
Bill Spight wrote:
Now let us add other simple gote similar to T and T1 to make an environment, such that T >= T1>= T2 >= . . . > 0

We still have three comparisons.

1) Black plays sente. Result: T - T1 + T2 - . . . .

2) Black plays to G. Result: G - T + T1 - T2 + . . . .

3) Black plays to T, then White plays to -R. Result: -R + T + T1 - T2 - . . . .

Comparing 2) to 3) we get that Black should not play to T if G + R > 2T. Same as before. :)

If 2T > G + R we compare 1) to 3) and get that Black should play sente if R > 2(T1 - T2 + . . .).

If G + R > 2T we compare 1) to 2) and get that Black should play gote if G > 2T - 2(T1 - T2 + . . .). I have separated T from T1, T2, etc., because T does not appear in the comparison between 1) and 3). It really should not be considered part of the environment.

We may estimate T1 - T2 + . . . as T1/2. The estimate does not affect the comparison between 2) and 3), but it does affect the others. Using the estimate we get

1) vs 3): Compare R to T1.

1) vs 2): Compare G to 2T - T1.


IIYC, your first steps of comparison clarify whether Black should start by taking T and you rely on T ~= T1 to suggest approximating conditions for the comparisons (1) to (3) and (2) to (3).


No. All I assume is that T >= T1.

Quote:
However, you completely lose me with your comparison (1) to (2); why do you compare G to 2T - T1? (1) reduces to T/2 and (2) is G - T/2. Comparing these two terms gives G >?< T.


Because T does not appear in all three comparisons, it is not really part of the environment. Therefore it does not appear in any approximations.

Quote:
Another aspect I have not completely understood yet is why we need not consider the strategy of Black taking T then White taking T1 for the sake of deciding whether Black should start by taking T.


You are right. :) When we add T2 we may need a fourth comparison,

G >?< T2 - . . . .

Edit: However, T2 and other gote were added to make an environment.

Bill Spight wrote:
Now let us add other simple gote similar to T and T1 to make an environment, such that T >= T1>= T2 >= . . . > 0

If we include the fourth comparison, then T1 is not part of the environment, as it does not appear in that comparison, but I had intended it to be the top play in the environment. That is why I had excluded the fourth comparison. (It has been a while, so I had forgotten that that is what I was doing. Obviously, I did not make that clear. :( )

Later edit: I have worked out what I was doing afresh, and only three comparisons are needed. It's late at night, so I'll explain later. :)

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Last edited by Bill Spight on Fri Dec 30, 2016 1:38 am, edited 1 time in total.
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Post #38 Posted: Fri Dec 30, 2016 12:19 am 
Judan

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Let me try to understand your study afresh. To ease my thinking, let me use partly different letters for the values.


Presume
- Black's local sente with Black's profit value F (where F is also the 'Follow-up's move value') and White's Move value (to reverse sente) M,
- a simple gote with move value T,
- an ideal environment with the largest move value (temperature) U,
- T >= U > 0.


We know the approximations
- starting in the ideal environment is worth U/2,
- starting in the combination of the simple gote and ideal environment, if we simply assume this enhanced ideal environment, is worth T/2.


We have the strategies

1) Black plays locally sente (answered locally by White), then Black starts in the enhanced ideal environment. The taken values are F - F + T/2 = T/2.

2) Black plays locally (gote in the local sente) but White starts in the enhanced ideal environment. The taken values are F - T/2.

3) Black takes the simple gote, White plays locally (reverse sente), Black starts in the ideal environment. The taken values are T - M + U/2.

4) Black takes the simple gote, White plays the largest move of the ideal environment. So far, the taken values are T - U. However, the local black sente is still available and we must judge conditions whether White's play is correct or reverse sente better.


First, we must clarify whether Black should start locally or in the enhanced ideal environment. Conditions arise from comparing (1) to (3) and comparing (2) to (3). I do not understand yet whether further conditions can arise from comparing (1), (2) or (3) to (4).

- Comparing (1) to (3): T/2 >?< T - M + U/2 <=> M >?< (T+U)/2. We compare M to (T+U)/2. The condition M > (T+U)/2 suggests Black's local start. (Let us postpone ambiguous cases for now.)

- Comparing (2) to (3): F - T/2 >?< T - M + U/2 <=> F + M >?< 3/2 T + U/2. We compare F + M to 3/2 T + U/2. The condition F + M > 3/2 T + U/2 suggests Black's local start.

- Since we have two conditions for Black's local start, we conclude it if M > (T+U)/2 OR F + M > 3/2 T + U/2.


Now, I have a few questions about your remarks.

- You say: "T does not appear in all three comparisons." Why then does T appear in the taken values of strategies (1), (2) and (3)?

- You suggest to exclude the simple gote of value T from the environment. I do not understand your motivation because including it permits the approximation T/2 for starting in the enhanced ideal environment.

- If we assume the environment with largest move value U to be ideal, i.e., have the approximation U/2 for starting in it, may we also use the approximation T/2 for the enhanced environment including the simple gote and call it ideal? And vice versa? We know nothing about the detailed value distribution in the envionment (other than decreasing values > 0). So I think that we may as well assume the approximations for both the environment and the enhanced environment and call both ideal. Right?

- How do we analyse strategy (4)? How about, after Black has taken T, look at the position afresh but then from White's view of starting?

- You write (symbols substituted for mine): "If F + M < 2T we compare (1) to (3)". How does this condition arise? Above, I explain how to derive the condition F + M >?< 3/2 T + U/2. You also write not to presume T ~= U. But... if we do presume this, we can derive your condition as an approximation. How do you derive your condition? How do you derive it without it being an approximation? Why can we use the condition as a requirement for then comparing (1) to (3)?

- You write: "If F + M > 2T we compare (1) to (2)". How does this condition arise? Above, I explain how to derive the condition F + M >?< 3/2 T + U/2. You also write not to presume T ~= U. But... if we do presume this, we can derive your condition as an approximation. How do you derive your condition? How do you derive it without it being an approximation? Why can we use the condition as a requirement for then comparing (1) to (2)?


Second, suppose we have identified that Black must start locally. Next, we have to clarify whether White should reply locally or play in the environment. We decide this by comparing (1) and (2): T/2 >?< F - T/2 <=> T >?< F. We compare T to F. White replies locally if T < F and replies in the enhanced environment by taking the simple gote with value T if T > F.


I still have my question why do you compare 2T - U to F now? Your answer "Because T does not appear in all three comparisons, it is not really part of the environment. Therefore it does not appear in any approximations." does not explain this to me. To start with, it does not explain it because I think, as above, T appears in all three comparisons. I have just explained how we get the comparison T to F. Although 2T - U ~= T for every reasonable environment in practice, where T ~= U, and we can simplify 2T - U as T, I still do not understand why you would need 2T - U here at all and how do you get it.

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Post #39 Posted: Fri Dec 30, 2016 1:35 am 
Honinbo

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Cher Robert,

I typically do not rely upon memory, but reconstruct things. It is late, so let me just briefly say that, while indeed you can add a fourth comparison with T2, it is not necessary, because the results of T - T1 (Black plays in T and White replies in T1) dominate the results of Black plays in S and G. That being the case, you only have to consider when Black plays in T and White replies in R, as White will not reply in T1 unless the Black play in T is best, anyway.

I'll flesh that out later.

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Post #40 Posted: Fri Dec 30, 2016 9:29 am 
Honinbo

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OK. Let's start afresh. :)

We have a local position, P = {{Big | 0}, G || -R}, along with a number of simple gote, {T | -T}, {T1 | -T1}, {T2 | -T2}, where G > 0, R > 0, and Big/2 > T >= T1 >= T2 >= . . . >= 0. Big, G, R, and all the Ts are numbers. Black has the move.

Case 1). Black plays in P to {Big | 0 }. Then, since Big > 2T, White will reply in P to 0. Then Black and White will alternate taking the largest gote.

Result (for Black): T - T1 + T2 - . . .

Case 2). Black plays in P to G. Then White and Black will alternate taking the largest gote.

Result: G - T + T1 - T2 + . . .

Case 3) Black takes the largest simple gote.

Case 3a) White then replies in P to -R. Then Black and White will alternate taking the largest gote.

Result: - R + T + T1 - T2 + . . .

Case 3b) White then takes the largest simple gote.

Case 3bi) Black plays in P to {Big | 0}.

Result: T - T1 + T2 - . . .

Note that this result is equal to the result in Case 1).

Case 3bii) Black plays in P to G.

Result: G + T - T1 - T2 + . . .

Note that this result is greater than or equal to the result in Case 2) by the amount 2(T - T1).

Thus, case 3b) is at least as good for Black as cases 1) and 2). It dominates them.

If also the result for case 3a is greater than or equal to the results for cases 1) and 2), then it is correct for Black to take the largest gote outside of P. Likewise, if the result for case 1) is greater than or equal to the results for cases 2) and 3a), then it is correct for Black to play in P to {Big | 0}, and if the result for case 2) is greater than or equal to the results for cases 1) and 3a), then it is correct for Black to play in P to G. The three comparisons to make are between the results for cases 1), 2), and 3a). :)

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