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 Post subject: Re: Sente, gote and endgame plays
Post #81 Posted: Sat May 27, 2017 1:34 pm 
Judan

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I need to read your message more carefully. For now, just one note: I do not mean "playing the sente in descending order of threats" but mean "playing the sente in descending order of sente move values" (the profit values of the opponent playing reverse sente).

With move value | follow-up move value annotation and the two local sentes 1|6 and 2|3, the sente player starts with 2|3 because 2>1. Sequences:

1) 3 - 3 + 6 - 6 = 0

2) 3 - 1 + 3 = 5

3) 6 - 6 + 3 - 3 = 0

4) 6 - 2 + 6 = 10

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 Post subject: Re: Sente, gote and endgame plays
Post #82 Posted: Sat May 27, 2017 2:02 pm 
Honinbo

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RobertJasiek wrote:
I need to read your message more carefully. For now, just one note: I do not mean "playing the sente in descending order of threats" but mean "playing the sente in descending order of sente move values" (the profit values of the opponent playing reverse sente).


Sorry. Condition 3 contradicts that. :) Given no kos, you should play the smaller sente if its threat is large enough.

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Post #83 Posted: Sun May 28, 2017 4:47 am 
Judan

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For local sentes a|b, r|s with a+2b > r+2s, a > r, 2s > 2b, correct start with a|b but outside your no-reading conditions, here is an example:

4|5, 1|6

1) 5 - 5 + 6 - 6 = 0

2) 5 - 1 + 5 = 9 mistake on move 2

3) 6 - 6 + 5 - 5 = 0

4) 6 - 4 + 6 = 8 mistake on move 2

*******************************************************************************************

You claim the following conjecture 0:

Suppose no kos in the game tree. There are the non-identical local sentes a|b, r|s of the starting sente player, whose correct start in a|b is at least as good as his start in r|s in the following non-exhaustive cases, without having to read, in a (possibly empty) environment T >= T1 >=...:

a) a > r AND 2b >= 2s

b) a = r AND 2b > 2s

c) a < r AND 2b > r + 2s


What is a proof, if you have already done it?

*******************************************************************************************

Let there be the environment T >= T1 >=...> 0 (or T = 0 if the environment is empty).
The sente player has the local sentes M0|F0, M1|F1,... (Mi are the sente move values, Fi are the gote follow-up move values, Mi < Fi is the local sente condition) with M0 >= M1 >=... The temperature is low: F0, F1,... >= T.

I made the following conjectures:

1) The starting sente player can, e.g., play all his local sentes in sente in the order M0|F0, M1|F1,...

2) The reverse sente player's start is determined by the comparison M0 to twice the alternating sum T - T1 +...

*******************************************************************************************

Counter-example for conjecture 1:

Local sentes 2|8, 3|4. Environment 3.5.

Profit sequences:

1) 8 - 8 + 4 - 4 + 3.5 = 3.5 correct start with 2|8

2) 8 - 8 + 3.5 - 3 = 0.5 mistake on move 3

3) 8 - 3 + 8 - 3.5 = 9.5 mistake on move 2

4) 8 - 3.5 + 8 - 3 = 9.5 mistake on move 2

5) 8 - 3.5 + 4 - 8 + 4 = 4.5 mistake on move 3

6) 4 - 4 + 8 - 8 + 3.5 = 3.5 mistake on move 2

7) 4 - 2 + 4 - 3.5 = 2.5 mistake on move 1

8) 3.5 - 3 + 8 - 8 = 0.5 mistake on move 1

9) 3.5 - 2 + 4 - 4 = 1.5 mistake on move 2

*******************************************************************************************

Nevertheless, it may be instructive to see my failed sketch of a proof of conjecture 1 and suggest where I made mistakes in it:

Iteratively, the sente player can start the next available local sente with the index k or play in the environment with its updated temperature T. At iteration step k for a sequence, if the opponent incurs a loss due to a positive net profit, he never gets a chance to compensate it at a later iteration step because none of the sequences below played to the opponent's turn has a negative net profit favouring the opponent. His playing of M_k+1, M_y or M_z as below prevents the sente player from later playing M_k+1 - M_k+1 = 0, or F_y - F_y = 0 and F_z - F_z = 0; this means that the opponent incurs a earlier loss by playing M_k+1, M_y or M_z instead of later neither gaining compensation nor losing anything from M_k+1 - M_k+1 = 0, or F_y - F_y = 0 and F_z - F_z = 0.


Special case: afterwards, zero local sentes remain:

1) F_k - F_k = 0.

2) F_k - T >(*2) 0 is the minimising opponent's mistake.

3) T... is dominated because of (*2).

(1) is correct, the sente player has started all local sente sequences and the proposition is proven.


Special case: afterwards, only one local sente remains:

1) F_k - F_k = 0.

2) F_k - M_k+1 >(*4) 0 is the minimising opponent's mistake.

3) F_k - T >(*2) 0 is the minimising opponent's mistake.

4) T... is dominated because of (*2).


Else regular case: afterwards, at least two local sentes remain:

Let y ≠ z be indices so that k < y, z.

1) F_k - F_k = 0.

2) F_k - M_y + F_k - M_z >(*4) 0 is the minimising opponent's mistake.

3) F_k - M_y + F_k - T >(*2)(*4) 0 is the minimising opponent's mistake.

4) F_k - T + F_k - M_y >(*2)(*4) 0 is the minimising opponent's mistake.

5) F_k - T + F_k - T1 ≥(*2) 0 is the minimising opponent's mistake or (1) is equally good so, without loss of generality, we assume (1).

6) T - M_k.

If F_k > T, this is the sente player's mistake because he does not follow the simple sente strategy by failing to prevent the opponent from taking M_k. If F_k = T, move 1 in (1) is equally good so, without loss of generality, we assume (1).

7) T - T1. is dominated by (1) to (5) because, after the sente player has started all local sente sequences, he starts playing the alternating sum of the environment.


Footnotes:

(*3) By the definition of local sente, we have M_k < F_k for each local sente with index k = 0, 1, 2...

(*4) F_k >(*3) M_k ≥(*1) M_y ≥(*1) M_z.

*******************************************************************************************

My failing sketch of a proof of conjecture 2 depends on conjecture 1:

Additional Presupposition:

We have the net profit S = 0 (expressed from the opponent's value perspective) of the opponent's remaining local sentes still available to him after any possible reverse sente play of the starting reverse sente player. (I use ∆ for the alternating sum starting at the denoted value.)

Conjecture 2 repeated:

If F0, F1, F2,... ≥ T, the reverse sente player starts
in the environment if 2∆T ≥ M0,
locally if 2∆T ≤ M0.

Failing sketch of a proof:

1) T - S - ∆T1 = ∆T.

2) M0 - S - ∆T = M0 - ∆T.

After move 1 in (1) or (2), apply conjecture 1. Comparison (1) ? (2) <=> ∆T ? M0 - ∆T <=> 2∆T ? M0. This implies conjecture 2 because the reverse sente player starts in the environment in (1) or locally in (2).

*******************************************************************************************

Now, this is really disappointing. We have the comparison in conjecture 2 for one local sente and low temperature. The question remains: Can conjecture 2 be proven differently or what counter-example can be shown?

*******************************************************************************************

My last hope for a general advice for several local sentes of one or both players possibly in an environment possibly with high temperature are these conjectures (no kos now or later):

3) For a player, playing the first local move in his local sente M0|F0 is least as good as playing the first local move in his local sente M1|F1 if M0 >= M1 AND F0 >= F1.

4) For a player, playing reverse sente in the opponent's local sente M0|F0 is least as good as playing reverse sente in the opponent's local sente M1|F1 if M0 >= M1 AND F0 >= F1.

Can they be proven or which counter-examples exist?

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 Post subject: Re: Sente, gote and endgame plays
Post #84 Posted: Sun May 28, 2017 7:30 am 
Honinbo

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RobertJasiek wrote:
For local sentes a|b, r|s with a+2b > r+2s, a > r, 2s > 2b, correct start with a|b but outside your no-reading conditions, here is an example:

4|5, 1|6

1) 5 - 5 + 6 - 6 = 0

2) 5 - 1 + 5 = 9 mistake on move 2

3) 6 - 6 + 5 - 5 = 0

4) 6 - 4 + 6 = 8 mistake on move 2



When those are the only plays on the board, as I said, a+2b >?< r+2s is a good guide, with no reading. 4 + 10 > 1 + 12, so start with A. Correct play gives the same result, but White's mistake is more costly. :) There is a psychological point, OC. White might be more likely to make the smaller mistake. ;)

Quote:
You claim the following conjecture 0:

Suppose no kos in the game tree. There are the non-identical local sentes a|b, r|s of the starting sente player, whose correct start in a|b is at least as good as his start in r|s in the following non-exhaustive cases, without having to read, in a (possibly empty) environment T >= T1 >=...:

a) a > r AND 2b >= 2s

b) a = r AND 2b > 2s

c) a < r AND 2b > r + 2s


What is a proof, if you have already done it?


I trust that a) and b) are obvious. :) Let's try c) with {2b|0} + {a||0|-2b} + {2s|0||-r} + {0|-2s}. White to play cannot win.

1) White plays in A, Black replies in R for jigo.
2) White plays in B, Black replies in R, White plays in A with sente. Result: Jigo.
3) White plays in S. Not! since 2b > 2s.
4) White plays in R. Black replies in B. White plays in A with sente, then plays in S. Result: 2b - r - 2s > 0. Black wins.


Quote:
My last hope for a general advice for several local sentes of one or both players possibly in an environment possibly with high temperature are these conjectures (no kos now or later)


I applaud your attempts to come up with general advice. :) I remember struggling with only 3 sente circa 1980. OC, that was long before I learned CGT. The CGT general results with only 2 sente leave a lot unsaid, and there are surely regularities with sente that do not meet the CGT criteria, along with other plays. The question is whether the rules are easy for humans to remember and how often the situations arise in actual games.

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 Post subject: Re: Sente, gote and endgame plays
Post #85 Posted: Mon May 29, 2017 2:58 am 
Judan

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Conjecture 3 (repeated)

In a position with possibly an environment and possibly several local sentes, there are local sentes M0|F0, M1|F1 with M0 > M1 and F0 > F1 of the starting sente player. Starting in M0|F0 is at least as good as starting in M1|F1.

Draft of a Proof

S is the starting sente player's strategy for (1) to (6). The first F0 is the first move. The last three denoted values of a bracket can occur in any order and be preceded, interrupted or succeeded by moves elsewhere, except that the second F1 is a follow-up and can only be played after the first F1 has already been played.

S' for (7) to (12) is S except for the denoted values. The first, second, third and fourth denoted value of S is substituted by the respective denoted value of S'.

We ignore the profit values of the plays elsewhere because replacing S by S' keeps them constant and does not affect the comparisons of net profits below.

1) S(F0;-F0,F1,-F1) = 0

2) S(F0;-F0,F1,F1) = 2F1

3) S(F0;-F0,-M1) = -M1

4) S(F0;F0,F1,-F1) = 2F0

5) S(F0;F0,F1,F1) = 2F0 + 2F1

6) S(F0;F0,-M1) = 2F0 - M1

7) S'(F1;-F1,F0,-F0) = 0

8) S'(F1;-F1,F0,F0) = 2F0

9) S'(F1;-F1,-M0) = -M0

10) S'(F1;F1,F0,-F0) = 2F1

11) S'(F1;F1,F0,F0) = 2F0 + 2F1

12) S'(F1;F1,-M0) = 2F1 - M0

These comparisons are constant: (1) = (7), (5) = (11). For the comparisons (3) > (9), (4) > (10), (6) > (12), the starting sente player incurs a loss by changing from S to S', that is, replacing the start with F0 by F1.

The comparison (2) < (8) is the starting sente player's gain. [---] Therefore, it remains to be shown that (2) does not occur due to strategic choices.

Can you either complete the proof or provide a counter-example?

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 Post subject: Re: Sente, gote and endgame plays
Post #86 Posted: Mon May 29, 2017 7:43 am 
Honinbo

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RobertJasiek wrote:
Conjecture 3 (repeated)

In a position with possibly an environment and possibly several local sentes, there are local sentes M0|F0, M1|F1 with M0 > M1 and F0 > F1 of the starting sente player. Starting in M0|F0 is at least as good as starting in M1|F1.

{snip}

Can you either complete the proof or provide a counter-example?


Using CGT notation, we are comparing playing first in {2*F0 | 0 || -M0} with {2*F1 | 0 || - M1}, plus or minus some constant, with F0 > M0 > M1 and F0 > F1 > M1. The no ko conditions in the environment apply.

To prove: Playing to {2*F0 | 0} + {2*F1 | 0 || -M1} is at least as good as playing to {2*F1| 0} + {2*F0 | 0 || -M0}. That is, D = {2*F0 | 0} + {2*F1 | 0 || -M1} - {2*F1 | 0} - {2*F0 | 0 || -M0} is at least as good as a score of 0. That is, even if White plays first in D, Black can reply to a score >= 0.

Proof:

Let A = {M0 || 0 | -2*F0}
Let B = {2*F0 | 0}
Let Y = {2*F1 | 0 || -M1}
Let Z = {0 | -2*F1}

Because F0 > F1, we already know that White will not start in Z.

Case 1: Let White start in A. Then Black will reply in Y, leaving

B - B + Z - Z = 0.

Case 2: Let White start in B. Then Black will reply in Y Leaving

A + Z - Z = A. We may ignore the miai, Z - Z.

Then White will play in A and Black will reply to 0.

Case 3: Let White start in Y. Then Black will reply in A, leaving

B + Z + M0 - M1.

Since F0 > F1, White will play in B and Black will reply in Z, leaving

M0 - M1 > 0.

QED. :)

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 Post subject: Re: Sente, gote and endgame plays
Post #87 Posted: Mon May 29, 2017 12:38 pm 
Judan

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Wonderful, very nice!

I confirm the proof of your lemma.

IIUC, your proof completes the proof of my conjecture 3, turning it into an important theorem. Right?

The details may be worked out: a careful description of the strategy; mentioning that each of your cases represents a "good enough" choice by Black on move 2 so that we do not need to consider his alternative move 2s.

The constant in your proof accounts the plays elsewhere, which I have assumed to be other local sentes or simple gotes of the environment. However, we may as well also allow local gotes with (even iterative) follow-ups as plays elsewhere and allow the other local sentes to have iterative follow-ups, if only the two speicific local sentes do not have iterative follow-ups (nor a follow-up to the reverse sente), there are no kos and all local endgames are separate to form a CGT sum. Right?

What happens if there are basic endgame kos elsewhere?

What do the theorem and its proof mean for my conjecture 4 of the starting reverse sente player?

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Post #88 Posted: Mon May 29, 2017 2:04 pm 
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RobertJasiek wrote:
Wonderful, very nice!

I confirm the proof of your lemma.

IIUC, your proof completes the proof of my conjecture 3, turning it into an important theorem. Right?

Right. :)

Quote:
The details may be worked out: a careful description of the strategy; mentioning that each of your cases represents a "good enough" choice by Black on move 2 so that we do not need to consider his alternative move 2s.

One of the nice things about difference games is that "good enough" is good enough. You don't necessarily have to find best play. :)

Quote:
The constant in your proof accounts the plays elsewhere, which I have assumed to be other local sentes or simple gotes of the environment. However, we may as well also allow local gotes with (even iterative) follow-ups as plays elsewhere and allow the other local sentes to have iterative follow-ups, if only the two speicific local sentes do not have iterative follow-ups (nor a follow-up to the reverse sente), there are no kos and all local endgames are separate to form a CGT sum. Right?

Right, except by constant I meant a number. I talked about the environment earlier. In this post I did not mention the environment explicitly, but in the difference game you get E - E = 0, where E is the environment. If there are kos in the environment or its tree, then we cannot say that E - E = 0, because kos are not combinatorial games. But if there are no kos, you can have anything in the environment. :)

Quote:
What happens if there are basic endgame kos elsewhere?

I suppose that you mean a ko with a swing of 1 pt. by territory scoring, 4 pts. by area scoring. As I have indicated in This 'n' That, we can treat such a ko as having a temperature as high as 0.5 by territory scoring or 2 by area scoring. If M0 > M1 > 0.5 by territory scoring or M0 > M1 > 2 by area scoring, I suppose that those sente will probably disappear before the ko is fought.

Quote:
What do the theorem and its proof mean for my conjecture 4 of the starting reverse sente player?

Your conjecture:
Quote:
4) For a player, playing reverse sente in the opponent's local sente M0|F0 is least as good as playing reverse sente in the opponent's local sente M1|F1 if M0 >= M1 AND F0 >= F1.


You can relax the conditions to allow F1 > F0. Playing in the reverse sente, {M0 || 0 | -2*FO} is at least as good as playing in the reverse sente, {M1 || 0 | -2*F1} if

M0 >= M1 AND M0 + 2*F0 >= M1 + 2*F1.

We can write the second condition as M0 - M1 >= 2*(F1 - F0)

So we would choose to play in {5 || 0 | -11} over {3 || 0 | -12}. :)

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 Post subject: Re: Sente, gote and endgame plays
Post #89 Posted: Sat Jun 03, 2017 2:20 am 
Judan

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Mathematical Go Endgames discusses move order of some shape classes of local endgames with iterative follow-ups if the initial moves have move values 1 or smaller.

What about positions with local endgames with iterative follow-ups and also larger move values?

1) What, if any, general simplification of reading is known (other than playing the environment of simple gotes in decreasing order)?

2) Can the theory of Mathematical Go Endgames be applied to, say, moves with move value 3 quite like it can be applied to moves with move value 1 (even if there are move values 3 and move values 1)?

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Post #90 Posted: Sun Jun 04, 2017 5:13 pm 
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RobertJasiek wrote:
Can the theory of Mathematical Go Endgames be applied to, say, moves with move value 3 quite like it can be applied to moves with move value 1 (even if there are move values 3 and move values 1)?


Sure. E. g., {15 | 6 || 3 ||| 0} vs. {6 | 0} is like {5 | 2 || 1 ||| 0} vs. {2 | 0}. :)

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

Quote:
What, if any, general simplification of reading is known (other than playing the environment of simple gotes in decreasing order)?


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

And I have done a good bit of work comparing games with variables instead of constants. I studied {a | b || c | d}, with numbers a ≥ b ≥ c ≥ d, back in the '70s. I have also done a good bit of work with difference games. :) Some of what I have found is fairly general. OTOH, some of those generalities strain the ability of humans to apply.

The idea of an environment of simple gote of the form, {a | b}, a ≥ b, is useful because we can compare other plays with simple gote of different sizes, and that comparison tells us, in general, when to play them. I don't know what use there might be for an environment of plays of the form, {a | b || c}, a ≥ b ≥ c, 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. So we could have such a well-ordered environment. :) (Note that the plays could be sente, gote, or ambiguous.)

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 Post subject: Re: Sente, gote and endgame plays
Post #91 Posted: Sun Jun 04, 2017 10:05 pm 
Judan

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Bill Spight wrote:
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...)

Quote:
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.

Quote:
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.

Quote:
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.

Quote:
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.

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Post #92 Posted: Mon Jun 05, 2017 7:22 am 
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RobertJasiek wrote:
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.

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Post #93 Posted: Mon Jun 05, 2017 7:50 am 
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RobertJasiek wrote:
Bill Spight wrote:
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.

Bill Spight wrote:
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.


RobertJasiek wrote:
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.

RobertJasiek wrote:
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. :roll: 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. :)

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Post #94 Posted: Mon Jun 05, 2017 8:53 am 
Judan

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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.

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Post #95 Posted: Mon Jun 05, 2017 10:35 am 
Honinbo

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RobertJasiek wrote:
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. :)

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Post #96 Posted: Mon Jul 17, 2017 8:37 am 
Judan

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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.

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Post #97 Posted: Mon Jul 17, 2017 12:24 pm 
Honinbo

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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:

Quote:
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.

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Post #98 Posted: Tue Jul 18, 2017 4:30 am 
Judan

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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:) )

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Post #99 Posted: Tue Jul 18, 2017 8:41 am 
Honinbo

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RobertJasiek wrote:
(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.

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Last edited by Bill Spight on Tue Jul 18, 2017 12:11 pm, edited 1 time in total.
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Post #100 Posted: Tue Jul 18, 2017 10:16 am 
Judan

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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.

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