Sente, gote and endgame plays

[Bill Spight]

$$B
$$----------------------------
$$|X X X X X X . . . . X . X .
$$|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 . .

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.

$$B Black's gote option, G = (GB + GW) / 2 = -33.5
$$----------------------------
$$|X X X X X X . . 2 3 X . X .
$$|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 . .

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.

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).

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.

[RobertJasiek]

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.)

[RobertJasiek]

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.

[Bill Spight]

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.

[RobertJasiek]

Quotation reference: http://www.lifein19x19.com/forum/viewto ... 24#p200624

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 http://www.lifein19x19.com/forum/viewto ... 50#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".

[RobertJasiek]

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.

[Bill Spight]

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.

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.

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.

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.

[RobertJasiek]

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.

[Bill Spight]

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.

[Bill Spight]

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).

[RobertJasiek]

We were talking about apples and oranges:) Now you clarify that you study a local sente with G as an alternative local option for Black and that so far you only study the case Big/2 > T, i.e., a special case of the follow-up move value aka Black's starting sente profit value being larger than T.

Thank you for your relevant research! I confirm the details of your message.

Your finding should be generalised a bit by not prescribing (all) the sizes of the leaves Ti, -Ti, Big, 0, maybe G and -R. Instead, use move values. Thereby you can describe a larger class of positions.

White to move, White's sente with White to move and White's sente with Black to move are "exercises for the reader";) I.e., if you do not provide that theory, I fear that I need to work it out some time.

My previous message is about my research in a different class of positions: a local sente WITHOUT local gote option at Black's start. (Same "exercises" ahead, OC.) No wonder that I got confused about some terms when they have not been applicable because you have studied something else:)

When I will have bitten myself through this, some other projects are a study of several local sentes of one player in an (ideal) environment and a study of several sentes of both players in an environment. Do you know how naive I were? I thought it was all easy but already the one sente studies are tougher than I imagined. Solving the early endgame (in parts) in general is not that easy after all. Uhm, I should have known:( I better do not mention simplifying assumptions, such as new influence having no value:) However, I do think that the classes of positions we are researching in are important ones for practical endgame application.

[RobertJasiek]

In a local sente with the initial position's count C and the sente follower's count S, we have C = S and accordingly the net profit 0 of the sente sequence. I have been wondering how this should be proven and have come up with a sketch of a proof using the assumption of a local sente being defined by its move value M being smaller than its follow-up move value F. However, then it occurred to me that C could be defined(!) as C := S. I guess that there could be more complicated proofs relying on min-max reasoning, temperature graphs and whatnot. If we forget about the combinatorial game theory origins and start introducing modern endgame theory afresh, which necessary definitions and which proofs are the best and why? I am not interested in subsets of local sentes, but want to discuss (simple) local sentes of arbitrary counts of its followers.

[Bill Spight]

In a local sente with the initial position's count C and the sente follower's count S, we have C = S and accordingly the net profit 0 of the sente sequence. I have been wondering how this should be proven and have come up with a sketch of a proof using the assumption of a local sente being defined by its move value M being smaller than its follow-up move value F. However, then it occurred to me that C could be defined(!) as C := S. I guess that there could be more complicated proofs relying on min-max reasoning, temperature graphs and whatnot. If we forget about the combinatorial game theory origins and start introducing modern endgame theory afresh, which necessary definitions and which proofs are the best and why? I am not interested in subsets of local sentes, but want to discuss (simple) local sentes of arbitrary counts of its followers.

Well, that equation antedates combinatorial game theory. It goes back in mathematics at least to the early 20th century and in go theory at least to the early 19th century. OC, it is possible to build theory entirely upon final scores, but counts are useful theoretical constructs. AFAIK, we do not have any records of the reasoning of go players of the 19th century or earlier about the equation, but it is not too difficult to show that evaluating the count of a sente position the same as the count of a gote position will lead you to play the reverse sente too early in many situations. For instance, consider this one point sente: {20 | 0 || -1}. Evaluating the count as 4.5 instead of 0 would suggest that White should play the reverse sente before this gote: {4 | -4}. While it is possible to construct a position where that is the case, it is almost always wrong.

it occurred to me that C could be defined(!) as C := S.

Well, that is basically what Conway did in his definition of thermography. But, harkening back to the early 20th century, the method of multiples shows that C -> S in the limit, as the number of multiples approaches infinity. And in my redefinition of thermography in terms of minimax play in Berlekamp's universal environment, you can show that C = S in such an environment, because it is wrong for White to play the reverse sente in that environment when the temperature of the environment is greater than 1, and when the temperature of the environment is 1, White is indifferent to playing the reverse sente or playing in the environment.

[RobertJasiek]

My sketch of a proof indeed also presumes some simplistic environment because you have trained us to think like that:) I have a problem with this approach though: it should be applicable not only in idealistic environments but always. Therefore, it seems that the definition approach is more according to my taste. A limes approach would be an overkill because I do not study infinitely many multiples. I mentioned a minmax approach because of your standard examples convincing(?) Bill Taylor of using sente counts instead of only gote counts but I am not sure whether that could be defined independently of environments, and how.

[Bill Spight]

My sketch of a proof indeed also presumes some simplistic environment because you have trained us to think like that:) I have a problem with this approach though: it should be applicable not only in idealistic environments but always.

The main value that I see for the non-terminal values for positions and plays is to provide a heuristic for selecting plays or candidate plays. The largest play is usually the best play. Now, in an ideal environment the largest play is always the best play. And the go board in a real game is so often close enough to the ideal environment that the largest play is the best play. Also, we know that drops in temperature may signal when the largest play is not the best play, because you want to get the last play before the temperature drops. Go players figured this out long before there was any such thing as combinatorial game theory. (OC, they did not talk about temperature, which was borrowed from CGT, but they had the general idea.) Go plays and positions are only partially ordered, so there is no reduction to any number that applies in all environments.