Sente, gote and endgame plays

[RobertJasiek]

the insights about CGT infinitesimals can help us to realize when non-infinitesimals act like infinitesimals.

Uhm, how? With which applications? (I am still learning infinitesimals...)

games and, hence, sums of games can be simplified by recognizing dominance and reversals, both of which may be analyzed using difference games.

We can take dominance, traversal (reversal is a misnomer by CGT people creating unnecessary confusion with [e.g., joseki] reversal and offering no hint about meaning, whereas traversal offers a hint: just traverse the skipped part of a tree) and equal options for granted.

I have also done a good bit of work with difference games. :) Some of what I have found is fairly general.

I guess. We all await your publication on (also) this topic.

I don't know what use there might be for an environment of plays of the form, {a | b || c}, a ≥ b ≥ c,

For the early endgame, an environment of simple gotes is good enough. For the (very) late endgame, we do not need additional complexity by introducing any environment at all.

but we may order those plays such that, given two such plays, A = {a | b || c} and D = {d | e || f}, if d - e ≥ a - b and e - f ≥ b - c, then each player does at least as well to play in D as in A, with the usual caveat about no kos.

Considering these two plays (possibly in an environment of unknown details), is this useful during the early endgame? It seems useful during the late endgame and related to our recent proofs. I need to study the relation between this and our recent study. If this is genuinely different, do you already have the proof available?

***

When asking for general insight about iterative follow-ups, I have had in mind first of all the late endgame and local endgames (also) with iteration deeper than 1 step. Apart from dominance, traversal, equal options and occasional use of infinitesimals, do we still know essentially nothing general other than reading (and playing simple gotes in decreasing order)? I ask because reading quickly explodes even for just a few local endgames with follow-ups.

Everywhere I see endless praise of professionals said to be playing near perfect (late) endgame, but what is the justification for this? Has anybody done systematic studies (not just for one game or two) of whether professional endgames are played correctly? Proving correctness can be hard easily. Can we do it for, say, the late endgame stage with "only" ca. 7 local endgames with (possibly) iterative follow-ups and move values larger than 1 (together with some simple gotes and local endgames with iterative follow-ups and move value 1 or smaller)? Is the praise of professional endgame play not just our excuse for not actually studying the late endgame carefully? Asked from the complementary view, if we know that professionals play perfect late endgame, what theory is it that we should be knowing and applying WRT iterative follow-ups of local endgames with move values larger than 1? Recall the evidence of Mathematical Go Endgames that they (and we) do not even play move values 1 correctly:) Instead of myths, I want real, proven, applicable theory.

[Bill Spight]

Everywhere I see endless praise of professionals said to be playing near perfect (late) endgame, but what is the justification for this?

That was true of top pros in the late 19th century, maybe earlier. They had time to work out nearly perfect play in the endgame, and they did so. That is not true with modern time limits, especially when games are finished in one day or less.

[Bill Spight]

the insights about CGT infinitesimals can help us to realize when non-infinitesimals act like infinitesimals.

Uhm, how? With which applications? (I am still learning infinitesimals...)

See, for example http://senseis.xmp.net/?EndgameProblem24 . None of the plays are infinitesimals, nor are their sizes equal. But they are close enough that correct play with infinitesimals ( * + ^ + *) indicates correct play in the problem.

but we may order those plays such that, given two such plays, A = {a | b || c} and D = {d | e || f}, if d - e ≥ a - b and e - f ≥ b - c, then each player does at least as well to play in D as in A, with the usual caveat about no kos.

Considering these two plays (possibly in an environment of unknown details), is this useful during the early endgame? It seems useful during the late endgame and related to our recent proofs. I need to study the relation between this and our recent study. If this is genuinely different, do you already have the proof available?

I think that we both agree that an environment of such plays would not be of much help. However, if A and D are both on the board, a not uncommon situation, we know aside from ko considerations, that with correct play each player will play D before A, and we can simplify our reading accordingly. This is easy to prove with difference games.

When asking for general insight about iterative follow-ups, I have had in mind first of all the late endgame and local endgames (also) with iteration deeper than 1 step. Apart from dominance, traversal, equal options and occasional use of infinitesimals, do we still know essentially nothing general other than reading (and playing simple gotes in decreasing order)? I ask because reading quickly explodes even for just a few local endgames with follow-ups.

I have worked on generalized corridors, plays of the form, {a | {b | { c | {d | ... }...}, a ≥ b ≥ c ≥ d ≥ ..., and their opposites. I have a solution for a simple gote vs. a generalized corridor of any length. But how much practical use is it? For instance, for White to decide whether to play in {x | y} or {a | {b | { c | {d | {e | f}}}}}, without considering other plays on the board, may take up to 14 comparisons. Really? Do we expect players to remember those comparisons just in case? Besides which, they will often fail, and you have to read things out, anyway.

OTOH, rules such as the one I gave for comparing A and D are useful. They are easy to remember and opportunities to use them arise frequently.

[RobertJasiek]

Having thought about your A + D setting, you are repeating in different annotation that the sente move value and follow-up move value of D are larger than of A. We have already proven this. New is your pointing out of application to local sentes, ambiguous, local gotes or two types occuring. I have checked my proof of strategy, your proof of comparing the two local endgames and my proof of starting to play to the "reverse sente" sides without follow-ups (with your conditions, which I happen to have found independently a few days earlier, buried in a proof with too restrictive assumptions on the environment, now corrected according to your generic environment without kos). The proofs go without any reference to the types of local endgames, so you are right about the broader scope of application. Wow! Now, this is really useful for simplifying reading to some extent. The proofs only rely on our assumptions stated earlier.

Let me define: D is the _larger_ and A is the _smaller_ local endgame with simple follow-up for the starting player. We shall write: D >= A. Note that the ordering is partial because not every two such local endgames have this (or the inverse) relation.

[Bill Spight]

Having thought about your A + D setting, you are repeating in different annotation that the sente move value and follow-up move value of D are larger than of A. We have already proven this. New is your pointing out of application to local sentes, ambiguous, local gotes or two types occuring. I have checked my proof of strategy, your proof of comparing the two local endgames and my proof of starting to play to the "reverse sente" sides without follow-ups (with your conditions, which I happen to have found independently a few days earlier, buried in a proof with too restrictive assumptions on the environment, now corrected according to your generic environment without kos). The proofs go without any reference to the types of local endgames, so you are right about the broader scope of application. Wow! Now, this is really useful for simplifying reading to some extent. The proofs only rely on our assumptions stated earlier.

Let me define: D is the _larger_ and A is the _smaller_ local endgame with simple follow-up for the starting player. We shall write: D >= A. Note that the ordering is partial because not every two such local endgames have this (or the inverse) relation.

I think that it would be helpful to use CGT definitions. D >= A iff White cannot win D - A if he plays first.

Edit: Maybe we could use some notation like D-black for a play in D by Black and D-white for a play in D by White. Then we could write: D-both >= A-both iff D-black >= A-black and D-white >= A-white.

[RobertJasiek]

Let there be two local endgames A|B and C|D in sente move value and follow-up gote move value annotation, in which a player, call him the 'creator', can create (make available) the follow-up. The opponent we shall call the 'preventer'. So the local endgame A|B has the sente move value A and the follow-up gote move value B; the other local endgame C|D has the sente move value C and the follow-up gote move value D. Let the creator start.

Bill has proven for A >= C and B >= D the creator's correct start is playing at A.

Let us study a more general case: A >= C and A + 2B >= C + 2D. For the starting preventer, we have proven the correct start of playing at A. Thus far we thought that for the starting creator Bill's proposition was our limitation of generalisation. I think though that we can extend generalisation to the more general case also for the starting creator.

For quite some study time, I could neither prove this nor find any counter-example with A >= C, B < D (so that it is outside the proven scope by Bill) and A + 2B >= C + 2D.

Now, I think I can prove it but I feel unsecure and do not really understand whether the following is a proof or goof. Please comment. Even better, if I goof, provide the desired counter-example.

Proposition: If A >= C and A + 2B >= C + 2D, a correct start of the creator is playing at A.

Proof:

There are these alternating sequences:

1) A - B - C - D

2) A - C - B

3) C - D - A - B

4) C - A - D

The results of (1) and (3) are equal.

Therefore, the decision between starting at A or C is the decision between (2) and (4) having, from the creator's value perspective, these resulting counts:

Count 2) -C - 2D

Count 4) -A - 2B

This amounts to a possible correct start of the creator at A if

-C - 2D >= -A - 2B <=> C + 2D <= A + 2B.

Without loss of generality, we can also assume A >= C.

These are the two assumptions in the proposition for a correct start at A.
QED.

[Bill Spight]

Your proof is good when the two sente are the only plays on the board. You need more for the more general case, when there are other (non-ko) plays.

Using CGT notation, let

R = {2B | 0 || -A}
S = {2D | 0 || -C}

The plays are sente, so

B > A
D > C

Also

A >= C and A + 2B >= C + 2D

Also let

D > B

else the already proven case where B >= D applies.

Then A > C

else if A = C then

2B >= 2D

which is false.

So we have

D > B > A > C

Given R and S:

There are these alternating sequences:

1) A - B - C - D

2) A - C - B

3) C - D - A - B

4) C - A - D

The results:

1) 0
2) 2B - C > 0
3) 0
4) 2D - A > 0

Which means that 2) and 4) are errors on White's part.

The conditions, A >= C and B >=D, hold up for correct play. OC, it is unrealistic and sometimes impractical to assume correct play, but if we can prove that our play is best, assuming correct play from the opponent, that information is important.

Counterexample where S is correct play by Black and R is not. To R and S add

X = {B | -B}
Y = {D | -D}

Lines of play:

1) R - Y - S - S , result = B - D < 0
2) S - X - R - R , result = D - B > 0

The play in R to {2B | 0} leaves it miai with {B | -B}, with a value of B, and the play in S to {2D | 0} leaves it miai with {D | -D}, with a value of D.

[RobertJasiek]

Many thanks for the class of elegant counter-examples for a non-empty environment and their nice proof! (Just for clarity, you use local sentes for them while using the sente move value did not mean to prescribe local sentes, but any counter-examples are good, even local sentes:) )

[Bill Spight]

(Just for clarity, you use local sentes for them while using the sente move value did not mean to prescribe local sentes, but any counter-examples are good, even local sentes:) )

Oh, I didn't get that. Thanks.

Well for the general case we have this:

Given

R = {2B | 0 || -A}
S = {2D | 0 || -C}

A, B, C, D non-negative

With Black to play, R dominates S in every non-ko environment

if 2B > C + 2D or

( A >= C and

(2B >= 2D or

(A + 2B >= C + 2D and

A > 2D

)))

A > 2D is impossible when R is local sente, given 2D > 2B.

[RobertJasiek]

I do not understand what you want to show because I am confused about which are the assumptions. Is your message self-contained or does it presume earlier assumptions? Is R dominates S an assumption, a conjecture, an informal summary of what or something else? Is given 2D > 2B an initial assumption or part of a proof? Is given C > A an initial assumption or part of a proof? I think I can decompress your case analysis text though. However, it would also help me to know what overall you are trying to study in your message.

[Bill Spight]

Sorry. Maybe this is clearer.

Given

R = {2B | 0 || -A}
S = {2D | 0 || -C}

A, B, C, D non-negative.

In every non-ko environment, when is a play in R by Black at least as good as a play in S, assuming correct play by White?

Case 1) 2B > C + 2D

Case 2) C + 2D >= 2B >= 2D and
A >= C

Case 3) 2D > 2B and
A + 2B >= C + 2D and
A > 2D

----
In case 3

A > 2D > 2B so R is gote.

My edit about C > A was the result of momentary confusion. Sorry. I am deleting it.

In case 3

A > C.

[RobertJasiek]

Case 1 is straightforward. Do you have proofs for cases 2 and 3, which also consider the arbitrary environment? Do you try to find all cases, think these are all and, if yes, what is the proof for case completeness?

[Bill Spight]

Start with the 0 game,

{2B | 0 || -A} + {A || 0 | -2B} +
{2D | 0 || -C} + {C || 0 | -2D} +
E - E

A, B, C, D are non-negative numbers, E is any non-ko environment.

Let White play first in {C || 0 | -2D} and let Black reply in {2B | 0 || -A}. If Black wins or gets jigo, then her play is at least as good as White's.

After the first two plays we have this game:

{2B | 0} + {A || 0 | -2B} +
{2D | 0 || -C} + {0 | -2D} +
E - E

which equals this game:

{2B | 0} + {A || 0 | -2B} +
{2D | 0 || -C} + {0 | -2D}

The non-ko environment drops out.

White to play. With correct play, under what conditions does Black get at least jigo?

----

The conditions can be expressed differently. I made them mutually exclusive. Here is another way of expressing them, where they are not mutually exclusive.

Case 1) 2B > C + 2D

Case 2) A >= C and
2B >= 2D

Case 3) A >= C and
A + 2B >= C + 2D and
A > 2D

[RobertJasiek]

So the structure of a proof starts with playing the difference game, whose advantage is that the environment is part of the input but drops out quickly so case study is restricted to the two local endgames. The sketch of your proof needs to be worked out. I suppose you have done it in your mind and maybe I could do it in writing now that I see the proof structure but I suppose I lack time during the few days before the European Go Congress and during it.

Does the completeness of the cases pop out naturally during working out the proof? Anyway, interesting, thank you!

I also need to understand where your class of counter-examples fits into the picture of your implied new proposition.

Although the cases are interesting in theory, we might say that they are already slightly more complicated than players would want to apply in practice. Would you agree?

[Bill Spight]

So the structure of a proof starts with playing the difference game, whose advantage is that the environment is part of the input but drops out quickly so case study is restricted to the two local endgames. The sketch of your proof needs to be worked out. I suppose you have done it in your mind and maybe I could do it in writing now that I see the proof structure but I suppose I lack time during the few days before the European Go Congress and during it.

Does the completeness of the cases pop out naturally during working out the proof? Anyway, interesting, thank you!

I also need to understand where your class of counter-examples fits into the picture of your implied new proposition.

Although the cases are interesting in theory, we might say that they are already slightly more complicated than players would want to apply in practice. Would you agree?

Well, even with only four small games the game tree is fairly large. However, it can be pruned pretty drastically. The logical simplification of the tree is not always straightforward. I think that it is necessary to make the comparisons tractable, but the resulting simplifications are not always intuitive. This and that. {shrug} And I don't do all this in my head. I have many notebooks full of this stuff.

I think that there is some value in making the cases mutually exclusive, but I am not sure. Doing so puts a bit of strain on memory.

One advantage is that you can show a kind of completeness of the cases. For instance:

Case 1) 2B > C + 2D

Case 2) C + 2D >= 2B >= 2D and
A >= C

Case 3) 2D > 2B and
A + 2B >= C + 2D and
A > 2D

The bolded conditions show that the full range of 2B in comparison with C + 2D and 2D is covered. But making the cases mutually exclusive in this way adds one condition to Case 2. {shrug}

As for application, Cases 1 and 2 are pretty obvious. Why case 3 works is not all that apparent. For one thing the condition, A > C, is not expressed, although it follows from the first two conditions. The second condition has a certain intuitive feel, but it may not be clear why it matters. Familiarity with the situation when White has the choice of plays will probably help. The third condition is not intuitive. It implies that {2B | 0 || -A} is a gote with a relatively small follow-up. So that is something. A fair amount of the time none of the cases will be satisfied, for either choice of plays, which means that the comparison has to be read out, which is what people tend to do anyway. I suspect that in practice a lot of people would check cases 1 and 2 and skip case 3.

Edit: The condition, A > 2D, was what allowed me to come up with the counterexample. That condition requires R to be gote, and in the counterexample it is sente.