Bill, I know you have not caught up with the earlier messages yet but I am struggling with another fundamentally important topic.

In the general case, how to recognise Black's 3-move traversal (CGT reversal)?





This game tree is represented by the counts of its positions. Move values are annotated by M followed by a letter. Positions are annotated by P followed by a letter.


Presuppositions:

1) MD, MV, MW, MX are small.

2) We study the case of PB being a simple sente or ambiguous.

The other case of PB being a simple gote needs to be studied later.


Known facts:

3) C = (D + W) / 2 // count of simple gote

4) MC = (D - W) / 2 // move value of simple gote

5) B = D // sente count by (2)

6) MB = X - D // sente move value by (2)


Black's 3-move traversal:

PA shall have Black's 3-move traversal. As such, PA is NOT a simple gote. This is expressed by the following conditions:

7) B - A < B - C <=> A > C // Black 1 gains less than White 2

8) (B - V) / 2 > C // A = (B - V) / 2 expressed as simple gote count; the condition means Black 1 gains less than White 2

Since PA shall have Black's 3-move traversal, PA is NOT a simple sente. This is expressed by the following condition:

9) C - V >= X - D // PA has at least the sente move value as PB; White 2 should (equality: need) not reply


Conjecture:

You have suggested that we have Black's 3-move traversal if the gote traversal move value MA is at most MC, as follows:

MA <= MC <=> // by definition of traversal move value, by (4)
(D - V) / 2 <= (D - W) / 2 <=>
V >= W


Attempt of proof:

(8) <=>
(B - V) / 2 > C <=> // by (5)
(D - V) / 2 > C <=> // by (3)
(D - V) / 2 > (D + W) / 2 <=>
-V > W


Here I am stuck. This is almost the opposite of the conjecture to be proven. What trivial mistake have I made?

Cher Robert,

Assuming that A reverses through C to D, then the incentive to move from A to D is D - A. The incentive to move from C to D is D - C. If the incentive to move from C to D is at least as great as the incentive to move from A to D, then Black should keep going if White plays from B to C. I.e., if D - C >= D - A, or equivalently, if A >= C.

Incentives are move values, but are not the same as temperatures. It is not true that if t(C) = t(A) that A reverses through C, but it does seem fairly obvious that if t(C) > t(A) that it does (for non-kos). Perhaps one way to prove that is that if t(C) > t(A) then D - C > D - A. Perhaps that can be proven by induction.

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

Visualize whirled peas.

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"It is not true that if t(C) = t(A) that A reverses through C": Indeed. I think I have found a few examples rejecting this. Playing the difference game yielded different results than comparing move values.

A few weeks ago, I have proven the proposition below and its analogue for White's 3-move sequence. In particular, it relates C ≤ A <=> MA ≤ MC, as we want. Furthermore, a hundred examples exhibit this relation. Some of the examples have more complicated trees with unsettled PE, an alternating gote or sente sequence for White's start, PB or PE as simple sente or long sequences resulting in series of such conditions. I do not have any counter-example.

We want C ≤ A <=> MA ≤ MC for the general case, but so far we cannot prove it. The proposition presumes settled PD and PE and gote traversal values of PA. Obviously, the requirements of PE can be relaxed to generalise the proposition with more research effort. E.g., we can allow privileges. However, real problem is: what is the most general case?

You have wanted to claim C ≤ A <=> MA ≤ MC in the general case, but you need to prove it. Do my proposition and proof below give you enough information for this purpose?

If other methods, such as comparing the opponent's branches or playing the difference game, are inapplicable, you have suggested MA ≤ MC as the always applicable condition for identifying or rejecting traversal of a 3-move sequence worth playing successively. Prove it! :) Otherwise, I ask again: how, in the general case, do we distinguish possible from impossible reversal (to start with, at least for alternating 3-move sequences)?



For proposition 10, the game tree represented by the counts is:



An 'unsettled' position corresponds to a combinatorial game that is not a number. Initially, the condition A = (D + E) / 2 is the tentative gote traversal count of PA. The theorem confirms that we have this gote traversal count and Black's 3-move traversal sequence because the conditions C ≤ A < D <=> C ≤ (D + E) / 2 < D in proposition 10 express that Black's alternating sequence is worth playing successively, that is, can be traversed. During Black's initial sente sequence, the count A of the initial position becomes the smaller or equal count C so Black can incur an initial loss. He compensates it by moving to the count D of his final gote follower because D is larger than A, which means his gain.

We have Black's alternating 3-move sequence PA - PB - PC.


Presuppositions

0) We have the combinatorial game PA := {PB|PE} := {PF|PC||PE} := {PF||PD|PG|||PE} with unsettled PA, PB, PC and settled PD, PE.
1) A := (D + E) / 2.
2) MA := (D - E) / 2.
3) C := (D + G) / 2.
4) MC := (D - G) / 2.


Proposition 10 [Black's 3-move gote]

G ≤ E <=> C ≤ A <=> MA ≤ MC and A < D.


Proof

Part I: "MA ≤ MC" <=> "G ≤ E <=> C ≤ A":

MC ≥ MA <=>(2)(4) (D - G) / 2 ≥ (D - E) / 2 <=> D - G ≥ D - E <=> -G ≥ -E <=> G ≤ E <=> G/2 ≤ E/2 <=> (D + G) / 2 ≤ (D + E) / 2 <=>(1)(3) C ≤ A.

Part II: "A < D":

PA unsettled(0) => D > E => D > (D + E) / 2 > E =>(1) D > A.


Remarks

(1) and (2) say that PA has the count A and move value MA so is Black's long gote with his 3-move gote sequence. (3) and (4) characterise PC as a simple gote. Note that MA = MC is allowed but we need not speak of it as an 'ambiguous' case.

When applying proposition 10 to an example, we check the presuppositions (0), (3) and (4) and whether therefore PC is a simple gote indeed, make the hypothesis of the count A, need not verify G ≤ E, verify C ≤ A to confirm the count A and calculate the move value MA. If the sequence to PD follows dominating options, we need not verify A < D because it follows from the already checked presuppositions (0) and (1). However, if we do not know whether the sequence to PD follows dominating options, we can determine this by verifying or refuting A < D.

Informally, we call A and MA in the presuppositions tentative values. The proposition confirms the values so, after verification of the conditions, we can remove the 'tentative' tag. The proposition and proof imply that the count A and move value MA of PA are well-defined.

Future research can make proposition 10 more flexible for PE if only it does not alter (1) and (2).

Just a couple of brief comments.



Right.



Sorry if I gave that impression. Maybe I misspoke some time or other, or made a comment assuming a particular situation. What I have believed, albeit without proof, is this.

C < A if t(C) > t(A) and t(B) > t(A) > t(D)

Edit: To be clear, A, B, C, and D are games.



I really have given the wrong impression! Difference games are always applicable, sans ko. Temperatures can be a guide, but only a guide.



Counts alone are not even enough to establish temperatures.

Also, I must say that, viewing this kind of tree, I would assume that t(A) > t(E), t(A) > t(F), t(A) > t(G), and t(A) > t(D), t(B) > t(F), t(B) > t(G), t(B) > t(D), t(C) > t(G), and t(C) > t(D), by implicature. Otherwise, more nodes may be relevant and should be shown.

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

Visualize whirled peas.

Everything with love. Stay safe.

This kind of comment always greatly helps me so that I need not decipher by context.



Good that it is clear now:)



What do you mean by this?



Yes, some such implied assumptions are assumed for what we might call a gote-encased alternating sequence. I have written down some for another proposition but have not figured out which of those move values comparing assumptions are minimally necessary (according to our current knowledge) and which others are implied from those to be stated explicitly.

However, future research can relax some of these assumptions or study related conjectures. E.g., for a tree with Black's and White's alternating 3-move sequences.

You like to annotate local temperatures t() whilst I like to study local move values. Are these the same or do local temperatures have additional meaning relevant for CGT study?

As we know, there is no one theory of move values. (And I am not referring to the commonly used deiri values, which are the source of much confusion.) However, temperatures match traditional miai values and O Meien's "absolute" values, so I usually prefer to use them. Berlekamp advises starting analysis with finding temperatures. They give you default or baseline lines of play.

Over the board, you can't just start drawing thermographic diagrams, and besides, players were able to derive miai values long before thermography was invented, through the use of mean territorial counts.

Consider the tree in question:



To my eye, A - G are nodes in the game tree, IOW, games. The leaves, D, E, F, and G, are either numbers or have low enough temperatures to ignore. In any event, they have mean territorial counts, m(D), m(E), m(F), and m(G), or we might write them in lower case, d, e, f, g, as we would numbers.

We may analyze the tree bottom up, tentatively assuming that each branch node is simple gote, unless we disprove that assumption.

Then we have m(C) = (d + g)/2. We also have t(C) = (d - g)/2.

Next we assume that B is gote and try the value, m'(B) = (f + m(C))/2. (The ' indicates a trial value.) To be consistent with that assumption, we must have m'(B) >= d. (If m'(B) = d, then B is ambiguous.) If m'(B) < d, then B is sente, and we set m(B) = d. Note that we do not have to look at t(C) to decide whether B is sente, but it is if t'(B) < t(C).

The fact that we can detect sente without looking at the temperatures means that those of us who figure out counts first do not have to figure out temperatures for each node in every tree.

If B is sente, then to proceed further up the tree we do not need either the mean counts or temperatures for anything below B. (Edit: That is so in this case because we have assumed that we can ignore the temperature of D.) So let's assume that B is gote. Then let us try m'(A) = (m(B) + e)/2. If m'(A) <= m(C) then A is gote or ambiguous and t(A) = m(A) - e. But suppose that m'(A) < m(C). At trap for the unwary is to assume that A is sente and set m(A) = m(C). Even experienced analysts have fallen into that trap. You also have to check for traversal, and try m'(A) = (d + e)/2 and t'(A) = (d - e)/2. If then t'(A) < t(B) and t'(A) <= t(C) then we have traversal. (OC, if t(A) = t(C) we may not have reversal. ) The point is that the play to B raises the local temperature and we should keep going until we have a stable temperature. (At least, if there are no kos.)

OC, you don't have to start with the working hypothesis that every move is gote, but it is a clear and thorough method.

Edit: BTW, note that if e > g and if B is gote we have a traversal.

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

Visualize whirled peas.

Everything with love. Stay safe.

Very useful! Now that I see it, everything fits. I am surprised about two aspects in your text "If then t'(A) < t(B) and t'(A) <= t(C) then we have traversal. (OC, if t(A) = t(C) we may not have reversal.)".

I expected the traversal condition t'(A) <= t(B) <= t(C) but you specify it as t'(A) <= t(B), t(C). That is, the comparison t(B) ? t(C) is immaterial. Surprise!

The other surprise is the necessity for t'(A) <= t(B). I thought t'(A) <= t(C) would be sufficient.

Can you please cast some additional light on both aspects to overcome my remaining doubts?

***

I think a better structure for the method is: 1) study B, 2) test A for long gote, else 3) test A for a case analysis of the type of B.

***

How about Black's alternating 5-move (or longer) sequence? Let me guess: Traversal is t'(A) <= t(A1), t(A2), t(A3), t(A4).

EDIT: (1) <-> (2).

Besides the theoretical questions in my previous message, let me apply your method to an example. Is my application correct?

The first / second values assigned to the letters are the counts (mean values) / move values (local temperatures).



We calculate the values of C:



We find that B is a simple sente and calculate its values:



We test whether A has a gote traversal with t'(A) = (d - v) / 2 = 5.5.

Since the conditions t'(A) <= t(B), t(C) <=> 5.5 <= 3, 6 are (partially) violated, A does NOT have a traversal.

We test whether A is a simple gote with t''(A) = (m(B) - v) / 2 = 5.5. Note that we derive this from m(B) and not from D. We find t''(A) > t(B) <=> 5.5 > 3 fulfilled. Therefore, A is a simple gote with m(A) = (m(B) + v) / 2 = 4.5 and t(A) = t''(A) = 5.5.

Usually, if Black starts at A, White does NOT immediately reply at B. Later, we expect White to continue, in which case Black will reply immediately. A is a simple gote with White's simple sente follow-up.

Now, let us start one move earlier:



We test whether P has a sente traversal with t'(P) = x - d = 3.

Since the conditions t'(P) <= t(A), t(B), t(C) <=> 3 <= 5.5, 3, 6 are fulfilled, P has a traversal and is White's long sente with m(P) = d = 10 and t(P) = t'(P) = 3.

If White starts at _____P_____, traversal to D is possible.

This is unlike Black starting at A, from which there is NO traversal. This is so because A has a larger local temperature than P and larger than (at least) one of the followers' local temperatures.

Cher Robert,

Traversal is your term. It seemed to me do be able to describe what I was talking about, the traversal of the game tree. If you object to that meaning I will avoid it.



This idea is easy to defend in terms of thermography, but it is something that I was aware of even before thermography was invented, because I explained local sente in terms of an environment of unknown plays. I.e., we regard this as sente because it carries a threat that is probably larger than other plays on the board. The same logic applies to traversals. Once we have raised the local temperature we keep going until the temperature drops. (Sans ko, OC. )



Well, if t(B) < t'(A) <= t(C) then B is sente, and m(B) = d. We don't have to look any further to find the mean territorial count of A and its miai value. It is true that for reversal we ask if A >= C, but that is different than comparing temperatures. We have ignored E and G below a certain temperature, but their details might affect the comparison of the games A and C. That is, we do not have enough information to answer that question definitively.

Edit: Reconsidering in terms of the left scaffold of A (see my next note), maybe we should consider some cases when B is sente to be one of traversal, even when t(B) < t(A).



As may be. You can start with the assumption of traversal, and only abandon it if doing so leads to a contradiction. I started with the assumption that every play is gote because it is easy to explain what to do when that assumption fails.



Yup. The basic assumption for calculation is that the hottest play is best.

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

Visualize whirled peas.

Everything with love. Stay safe.

In our example tree, we are looking for the mean territorial count and temperature of A. They are determined by the intersection of their right and left scaffolds. Above that intersection, at higher temperatures, at least one player will (by assumption) have a better play elsewhere.

Above some temperature the right scaffold of A may be represented schematically as a left leaning line:



I.e., v = e + t

The left scaffold of a may be represented schematically as



I.e., v = max(d - t, min(m(C), m(B) - t))

If the intersections of the two scaffolds occurs on the bottom segment of the left scaffold, then we have traversal. The temperature of the intersection, t = (d - e)/2, must satisfy the inequalities, t <= d - m(C) and t < m(B) - m(C). If the intersection occurs in the middle segment, then A is sente, and if it occurs in the top segment, then A is gote without traversal.

If B is sente, then the left scaffold of A is simply the line, v = d - t, and we do not care, as far as calculation is concerned, with whether there is traversal or not. We may care about reversal, but then the details of E and G may be relevant.

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

Visualize whirled peas.

Everything with love. Stay safe.

As a go term, I use it for a particular alternating sequence consisting of 3+ moves and for a specific involved move as the longest such sequence worth playing successively locally in a local endgame. E.g., a local hane-connect sequence with large enough threats is a traversal sequence consisting of 4 moves. Only 2 moves of it I do not call so because it is not the longest such sequence and would have fewer than 3 moves. I would say that there are two concatenated parts together forming the traversal sequence.

If German to English translation applies, traversal of a graph has a very different meaning in maths / informatics.

***

Concerning your thermographic remarks, maybe they will be useful for next year. Currently, time restrictions force me to postpone thermography.

Can you please explicitly confirm or deny (then with explanation) - without referring to thermography - whether application of your method to my example is correct? I have studied evaluation of long sequences for 6 months now and still wonder whether finally I might have had my heureka moment of, what is like, deciphering linear B or whether I am still poking around in the dark.

You must understand my problem. I have read 1000s of text pieces and examples (most from you) in order to make sense of it. For lots of things, I could find no more than a subclause in a comment for a well hidden example. So I have needed very much guesswork, testing and mathematical research of my own to possibly reveal for what you seem to have at least a structural (and partly deeper) understanding. Explicit, direct comments on my understanding can help me a lot because I cannot take time to (dis)prove all my ideas as theorems.

The calculations are correct.



As I have mentioned, I have reconsidered the question of traversal. If there are no kos, then the paths from both P and A traverse to D. However, if there is a ko or possible ko, then in general we wish to preserve potential ko threats. Thus, with the move from A to B there is a drop in the size of a play from 5.5 to 3, White may wish to preserve the ko threat and not continue to C. But if there is no ko or possible ko, White should continue to C, with a traversal to D, to prevent Black from possibly playing from B to Y and gaining 3 pts. (Except, perhaps, for psychological reasons. )

OTOH, if White plays from P with sente, then White is willing to relinquish a ko threat to prevent Black from playing to a position worth 13 pts., so why not keep going from B?

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

Visualize whirled peas.

Everything with love. Stay safe.

Hm. Now I am happy to understand value calculations and unhappy because we must also consider the strategic context...

Bill, checking the conditions t(A) <= t(A1), t(A2)... against examples, I am running into the same problem that must have prevented me before from using the condition t(A) <= t(A1) for gote traversal sequences. Again, I think that the condition is best omitted. (For long sente sequences, we use it.)



C is White's simple sente. B is White's simple sente.

A must be Black's long reverse sente with his long gote sequence and White's short sente sequence.

Tentatively, t'(A) = d - f = 7 - 0 = 7.

This is consistent with White's alternating sequence: t'(A) < t(E) <=> 7 < 16.

This should also be consistent with Black's alternating sequence: t'(A) <= t(B), t(C).

Whilst t'(A) <= t(C) <=> 7 <= 7 is fulfilled, t'(A) <= t(B) <=> 7 <= 1 is violated!

If we omit the condition t'(A) <= t(B), we do not have any problem and have t(A) = t'(A).

m(A) = f = 0.



Black's profit from A to B is 7. White's profit from B to C is 7. After Black's move to B, White can reply to C because of constant profits. In other words, m(A) = m(C). We must not prevent White from replying by using the superfluous condition t'(A) <= t(B). This is so not only for long reverse sente but also for long gote (Black's long gote sequence, White's short gote sequence). Do you agree? Compare reversal: A and C are compared, regardless of B.

In CGT terms, A = C. So of course Black's play from A to B reverses through C to D.

However, the play from A to B gains 7 pts., making it a 7 pt. reverse sente. But B is a 1 pt. reverse sente. As a practical matter, in go it is normally right for White to stop at B, leaving the play from B to C as a ko threat.

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

Visualize whirled peas.

Everything with love. Stay safe.

We learn that it can be correct to calculate traversal values but this does not necessarily imply traversal play, although it is optional if ko threats need not be preserved. Or it can be correct not to calculate traversal values but traversal play can be possible. Calculation and play are two different things. Reversal is a third animal. There are relations but they depend on additional value conditions. Maya script!

Maybe I should overcome my prejudice and perceive the example as an ordinary reverse sente with Black's 1-move gote sequence followed by White's sente follow-up. Initial values are the same.

I think that the 1 is a typo for 7.

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

Visualize whirled peas.

Everything with love. Stay safe.

To a first approximation, average counts and miai values are a good guide. For more accuracy, further thermographic information may be useful. Often ko considerations apply. And sometimes difference games can cut through uncertainties and ambiguities.

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

Visualize whirled peas.

Everything with love. Stay safe.

Without smiley, I might suggest to override move values by profit values of the currently next moving player during the alternating sequence. Then the 1 is overridden by the 7. Not that I would have worked out a complete theory for such, shall we say, incentives. Traversal is a matter of interpretation, isn't it?