It is time-consumimg, especially when writing. I have written hundreds accurate calculations with verifications and drafted many more but do not expect me to do everybody else's work. It takes seconds to show a difficult-to-calculate position but days to write down its accurate solution. You know where you find my explanations of how to calculate and verify accurately (ignoring infinitesimals).

Yes Robert I understand using the pure theoritical approach to calculate the miai of such position is far too difficult in practice.

Searching another approach I have a pure theoritical question:

Let's take the above tree : nodes C and E are supposed to be leaf nodes. Taking leaf C as the reference let's assume the count of position C is "0" and the count of position E is "e".
Now let's suppose you do not know the subtree under node D but let's suppose you know for sure that the miai value of node D is egal to vd.
My question is the following one. Can we conclude that the miai value va of node A is :
va = min(e, e/2 + vd/2)

If not can you show a counter example?

Oops, it is not exactly the tree I had in mind. Wait a little till I correct this tree. Sorry.

After correction:

Let's take the above tree : nodes C and E are supposed to be leaf nodes. Taking leaf C as the reference let's assume the count of position C is "0" and the count of position E is "e".
Now let's suppose you do not know the subtree under node B but let's suppose you know for sure that the miai value of node B is egal to vb.

My question is the following one. Can we conclude that the miai value va of node A is :
va = min(e, e/2 + vb/2)

If not can you show a counter example?

No, this is your exaggerating misreprentation. It is often not too difficult but it quickly becomes time-consuming, and writing down a practical execution of the theory applied to a difficult-to-calculate example is much more time-consuming.



You have fallen back into your habit of swapping the Black = Left convention for trees; it must be max instead of min.

I am not amused by your writing vd instead of d, unnecessarily complicating matters.

Anyway, I will await your correction.

Some things that one might have to check is tree simplication theorems / methods and whether ko might disturb the task.



No time.

[quote="RobertJasiek"]

You have fallen back into your habit of swapping the Black = Left convention for trees; it must be max instead of min.

I am not amused by your writing vd instead of d, unnecessarily complicating matters.[quote]

Why do you suppose I did not follow the convention Black = left?
I used "vd" to avoid ambiguity between the counts a, b, c, d .. and the miai values va, vb, vc, vd. OC I am open to any other convention.

I expected different things, sorry!

Please explain what you calculate here: min(e, e/2 + vb/2)


Instead of vd, I write Md, where M stands for move value, you know:) Easier! Actually I write D with capital letter as an index, which is hard here in text discussion. Bill preferred lower cases of maths annotation. I prefer upper case as easier to read amidst prose.

As mentionned in https://lifein19x19.com/viewtopic.php?p=277558#p277558, in my mind, the value min(e, e/2 + vb/2) is expected to be the miai value of A position.
The value "e" correspond to a sente black move (reverse sente white move) and the value e/2 + vb/2 to a gote move (black and white)

I get your notation now but have not understood why you take the min. Even when I know, I do not think that I can simply answer your question. If I suspect such a condition, I try to prove it with algebra. Maybe you need CGT simplification of the type reversion. Anyway, beware of kos due to the subtree, whose impact I cannot exclude (yet).

OK let's take two examples (the first is a gote situation, the second a a (reverse) sente situation


OC, in the general case I consider here the node "?" may have a large subtree but I did not find a situation where my simple calculation fails. That is the point where you can surely help.

Your example was confusing to me for the reason that it appears that it sometimes won't work unless C = 0 and all the values are positive. I think I know what you mean, it's the following kind of thing. I'm not sure why you suspect there is a counter example



(Black moves go to left with positive numbers and white moves go to right with negative numbers.)

This basically follows the pattern that the move value is the minimum of the value assuming sente and the value assuming gote.

We can write
E + T(B) - T(A) = C + T(A)
T(A) = (E - C) / 2 + T(B) / 2
for gote moves and
E = C + T(A)
T(A) = E - C
for sente moves. These are relationships that can be read directly from the game tree above. Now, when a player makes a move it's the other player's choice if to defend in gote or allow the follow up. That is, it's up to the other player if we get the sente or gote case. It follows that the move value is the lesser of the gote and sente since it is the second player's choice and the lesser is the better choice for that player.

Btw if it was not clear why (E - C) / 2 + T(B) / 2 is the gote move value then to see this note that
B = E + T(B)
and therefore
(B - C) / 2 = (E + T(B) - C) / 2 = (E - C) / 2 + T(B) / 2

The first time I heard about this was about 10-15 years ago, a Japanese friend was taking some lessons from a Japanese professional player who thought him something like this, not as pseudo mathematics though, at least this where he said he had this knowledge from.

Hope I didn't make too many typos and thinkos.

kvasir, such algebra is a means to study these aspects indeed. You write T for local temperature, which is our simple games is the move value.



Here I am lost.



For Black's sente moves! :)



We can for simple local endgames with settled D, E, F, G.

For Gerard's partially more general case, we can start with the assumption of no kos. For the general left subtree D, long alternating traveral sequences starting from the initial position A do not occur. However, when assuming known T(D), short or long alternating traveral sequences starting from D can occur (without kos by assumption), or at D Black's correct move might be a pass. I have not reflected (yet) whether the subtree might have to be studied or the mere assumption of known T(D) is sufficient but this is what you need to study, Gerard.

CGT reversal alone cannot simplify the subtree in general. The method of "comparing the opponent's branches" is another candidate to possibly simplify the subtree. However, even such a method a) considers the left alternating follower X of D and b) compares E >= E'', where E and E'' are the positions and E'' is the position two alternating moves succeeding E.

Therefore, I do not think that you can just write '?' for the children of D but you need to assess at least X, E and E'', unless you come up with yet other general tree simplification methods.



Some such techniques have been known subconsciously by a few users of traditional endgame theory but I have never seen them described explicitly. In particular, Richard Bozulich has applied something like this for three problems but I had to infer his doing and the presumed value manipulation technique by myself, which I only could because I had already done algebra of your type (the type of adding / subtracting [half] a gote move value to travel to / from a grandchild).

Oops, changing tree may lead to completely different conclusion. To understand what I mean let's take another very different tree:

With this tree even if you know the count of c and d of positions C and D and the miai value vb of B you cannot deduce the miai value of A as proved by the two following examples with c = 0, d = 12 and vb = 5:

To answer my question you are only allowed to build a subtree of node D, nothing else.

Note : I agree with Kvasir showing that a "min" rather than a "max" has to be used. Thank you for your help.

Yes Robert long alternating traveral sequences starting from the initial position A cannot occur. That is a key point of my tree proposal.

I do not understand what you mean Robert because for me E is assumed to be a leaf.

I spoke about E in a more general context but are fine with your different study scope of settled E. However, settled E has the consequence that we cannot apply the method of comparing the opponent's branches to possibly simplify the subtree of D. By setting E settled, you are not necessarily easing your task (yet) because this way of simplifcation is unavailable:)

Let's try this reasonning:

if va < vb then black AB move is sente and white have to answer with BE => va = e (e - c with c = 0)
if va > vb then:
White to play => white plays AC and black play in the environment => score1 = 0 + t = t
Black to play => black plays AB gote and white plays in the environment. Then, when temperature of the environment drops to vb, because vb is the miai value of B we can assume that white will play BE and black will play in the environment to gain vb points => score2 = -t + vb + e
you can now calculate va:
score1 = score2 => t = -t + vb + e => 2t = (vb + e) / 2
and va = vb/2 + e/2

Eventually you can conclude that va = min (e, e/2 + vb/2)

With this reasonning I do not mind if a ko exist in the subtree of B. I only assume that the environment is ideal.

I will amend as follows to be clearer.



You can derive all of these formulas by writing down relationships between the nodes.

For example for D and C. It's possible to do this in different ways but if you start with these three relationships

D - vb = B
B - va = A
A = va + C

Which are true when we talk about the gote (unforced play) case, they are not true if play is forced in these nodes.

It is fairly easy to go from this to
D - vb - va = va + C
at which point solve for va
va = (D - C) / 2 - vb / 2
The right side of which is the gote case in the "min" function of the third formula.


These two cases were shown


Here we need the correct value of vb to proceed to calculate va, if we have the correct value there is no problem.


First case:

Here we must be careful to work with the right temperature in the subtrees, here vb = 6 and ve = 8 since play is forced at lower temperatures. Now marking paths with * that are excluded for the reason that the other player's move is forced, we have this tree.

Using D = 12, C = 0, vb = 6, we can apply the formula

va = min(D - C, (D - C) / 2 - vb / 2)
= min(12 - 0, (12 - 0) / 2 - 6 / 2)
= min(12, 6 - 3) = 3


Consider the subtree

If we make the error to not realize that play is forced then we get incorrect values, that is values that could be correct if this was not a follow up from a lower temperature position.

Using B = 12, F = 6, vc = 4, we can apply the formula

va = min(B - F, (B - F) / 2 + vc / 2)
= min(12 - 6, (12 - 6) / 2 + 4 / 2)
= min(6, 3 + 2) = 5

or using B = 12, G = -2, vc = 4, we can apply the other formula

va = min(B - G, (B - G) / 2 + vc / 2)
= min(12 - (-2), (12 - (-2)) / 2 - 4 / 2)
= min(14, 7 - 2) = 5

If we go ahead to use this value in the previous case we get an incorrect result

Using D = 12, C = 0, vb = 5 (wrong!!), we can apply the formula

va = min(D - C, (D - C) / 2 - vb / 2)
= min(12 - 0, (12 - 0) / 2 - 5 / 2)
= min(12, 6 - 2.5) = 3.5

which is not the solution.


Second case:

Here too we must consider that play is forced at low temperature.

Using D = 12, C = 0, vb = 8, we can apply the formula

va = min(D - C, (D - C) / 2 - vb / 2)
= min(12 - 0, (12 - 0) / 2 - 8 / 2)
= min(12, 6 - 4) = 2



In this case there is something interesting, if we assume the following tree instead, we get a "correct" value (at least given how I filled in the tree at the top level) but it's not the value that I think we want.

Using D = 12, C = 0, vb = 10, we can apply the formula

va = min(D - C, (D - C) / 2 - vb / 2)
= min(12 - 0, (12 - 0) / 2 - 10 / 2)
= min(12, 6 - 5) = 1

Here we botched the calculation (again?). The error is that the first tree is better for black and black can force the result, that is black will play in gote as shown in the first tree.

Conclusion
I think this must work. You need to use the correct values for the position values and move values, that is you need to somehow have the right idea of which moves are forced and which are not. That means if you have two continuations that you can estimate the position value and can recall (or guess) a move value for the third hard to fathom continuation you can try to make a good estimate without working through the difficult continuation.

At least, seeing your post, it is obvious that we have the same understanding: applying similar formula to the two different examples above lead to difficulties due to gote-sente relationship.
From that point we have indeed two different approaches:
1) my approach is to identify on which kind of positions the formula can be easily applied, hoping that in practice, a majority of encountered positions are concerned
2) you approach seems (?) to adapt the miai value definition in such a way that the formula applies to (almost?) all kind of position.

General comment : if I understand correctly the theory the miai value is defined for a position ONLY when considering this position as the root of the analysis : looking at thermography, various scafolds are calculated recursively till the root and eventually, only on the root, the miai value is defined as the basis of the mast.
In my approach, when I am using the terme "miai value" I always consider the considered position as the root of the tree analysed.
On the other hand, my understanding of your approach is that you use a miai value (like vb or ve) in the subtree and this miai value depends on the above part of the tree in order to determine whether or not a branch is sente. If it is true then what is your definition of such intermediate miai value? I do not see clearly how you can decide that a branch is sente. If you know in advance the value va then you may know if the branch of a subtree is sente or not but this value va will be known only at the end of the process. For that reason I do not really understand how do you proceed.

Let me show you how you can use the result above to calculate the miai value of the following (simple) corridor

Click Here To Show Diagram Code

[go]$$
$$ -------------------------
$$ | X X . X X . X X . X . .
$$ | X X O X X O . O O X . .
$$ | X . O O O O O O . X . .
$$ | X . O . . . . . . . . .
$$ | O O O . . . . . . . . .
$$ | . . . . . . . . . . . .
$$ | . . . . . . . . . . . .
$$ -------------------------[/go]

The corresponding tree is the following:

You see that when black plays three times locally (and white plays three tenuki) then black gains 6 + 9 + 14

What is the miai value of the initial position A?
If you apply the thermography process then you have to calculate recursively various scafolds till the root.
Instead of scafolds you may also, for such simple corridor, calculate recursively the count, the miai value and perhaps the gain of each move.

My point is that you have OC to evaluate each leaf and, for intermediate nodes, you have only to calculate recursively the miai value of the node and nothing else.
Let's procede:
Firstly you calculate vd miai value : obviously vd = 7
Secondly you calculate vb miai value : 7 (vd) < 9 (g-e) => B is a gote position => vb = (7+9)/2 => vb = 8
Thirdly you calculate va miai value : 8 (vb) > 6 (e-c) => A is black sente => va = 6

How can you be more efficient?

Ok, but you need to clarify which kind of ideal environment you presume. Mine? Bills? Or rather Berlekamp's rich environment?



Ok.



For Black to gain t in the environment, you are making some implicit assumption(s). If you used my ideal environment, starting in it were worth t/2. Since you account t, you might have a different study approach, such as "White makes the first move and Black makes the last move in each variation". Or maybe you consider both players' starts but demand an equal number of moves or whatnot. Please clarify!



Ok.



Why can we assume a) White continues locally and b) B is a local gote?



You wish. However, how do we know that B is not Black's simple sente, Black's long gote, Black's long sente or a ko?



You assume the miai value (usually: the move value) vb of B. Why would this necessarily imply that White's gain on BE is vb? It would be vb if B were a simple gote, but see above (we do not know yet that it is).



You add the gains -t + vb and then add the count e. Sure, if the gains are right, we may do such.



You start by studying the equality score1 = score2. Not doing your work, please explain what is the purpose of considering this equality and why does it express this purpose?

t = -t + vb + e <=> 2t = vb + e <=> t = (vb + e) / 2 and not, as you write (accidentally?) 2t = (vb + e) / 2.



Please explain both steps!

(I have not read the next messages yet.)

I do not know in detail the definition of the various ideal environment you mentionned but I think my reasonning works with any of these environments

As usual I consider both players' starts but demand an equal number of moves or whatnot.

When temperature of the environment is greater than vb then it is correct for black to play in the environement (it may happen that a local black move could also be correct but playing in the environment is always correct).
When temperature of the environment is lesser than vb then the only correct move for black is to play locally (playing in the environment is not correct)
When temperature of the environment is equal to vb the situation is ambiguous and it is correct to play either in the environment or locally.
Note : the situation for white is exactly the same.

As a consequence assume temperature drops to vb.
If it is black to play it is correct to play in the environement and it is correct for white to answer locally
If it is white to play it is correct to play locally and it is correct for black to answer in the environment.

In any case I can always assume that white will play locally while black will play in the environement.

It does not matter if B is a local gote because I can assume that white will play locally while black will play in the environment

Here again it doesn't matter because black will not play locally

I never said that White's gain on BE is vb. I only said that black will gain vb points by playing in the environment (when temperature of the environment has dropped to vb)

Thank you for the correction of my typo in the formula.
On many occasion I use this score1 = score2 to calculate a miai value and you never argue on this point. What is wrong here?