Cards or app for miai-value based endgame practice?

[Gérard TAILLE]

Counter-example 1:

The problem of the method making a hypothesis is that, at low temperature T = 1,5, both options C and D are fulfilled, for option C giving the initial count -3 and move value 1 but for option D giving the initial count -2,5 and move value 1,5. A contradiction arises, which does not arise for thermography and the initial count -3 and move value 1.

Counter-example 2:

Method of making a hypthesis: initial count -1 and move value 2 derived from a 5-move sequence.

Thermography: initial count 0 and move value 3. Note that my description of thermography speaks of move value and I dare to do so because I ignore infinitesimals. I suspect that miai values with infinitesimals might differ from move values, but I do not know examples.

Interesting information.
As expected my method leads to same result as miai value.
Example 1 : I understand you reached a contradiction with method making a hypothesis => an improvment of the method has to occur.
Example 2 : I understand that you found a 5-move sequence. If it were true then I would agree that initial count = -1 and move value = 2.
In my method as well as in thermography the result is initial count 0 and move value 3.
I think it is wrong to consider the 5-move sequence PQ-QR-RS-SU-UV (with V being the leave with the count +1). This 5-move sequence could exist if temperature were under 1 but this sequence does is not corret when temperature is greater than 1. OK for the sequence PQ-QR when temperature drops under 3 but the answer RS cannot be done if temperature is greater than 1. That is where your method probably failed.
Do you really think that RS can be played when temperature is greater than 1?

[RobertJasiek]

a contradiction with method making a hypothesis => an improvment of the method has to occur.

The improvement is thermography. The problem with the method making a hypothesis is that it can sometimes produce wrong values, although no such example on the board is known yet. The problem with thermography is that it is inapplicable in practice during a game. Both methods have their merits and limitations.

In my method as well as in thermography

You speak about your method as if you applied it correctly but you do not calculate enriched scores and thermographs yet, as you should because your method is a sort of informal thermography.

Do you really think that RS can be played when temperature is greater than 1?

This is not the question. The question is: if the start is at P, does play traverse to U?

[Gérard TAILLE]

Do you really think that RS can be played when temperature is greater than 1?

This is not the question. The question is: if the start is at P, does play traverse to U?

It is the question: you say that the “move value” of P is equal to 2 => you start at P with temperature higher than 1 => RS cannot be played at this temperature => the 5-move sequence cannot be correct at this temperature.

[RobertJasiek]

You throw in temperature is a value to compare with, but then you must be specific: which temperature?

Bill and others compared the tentative initial move value to the follow-up move values but this was wrong, as my counter-example shows. I compare the tentative initial move value to the moves' gains, which is less but still wrong, as Francisco's thermography counter-examples show. I guess comparing the tentative initial move value to some temperatures will be better because it is non-infinitesimal thermography and might work as long as we do not consider infinitesimals. However, to apply that, we must use thermographic calculus (or graphs) - not just your wishful guessing.

[Gérard TAILLE]

Bill and others compared the tentative initial move value to the follow-up move values but this was wrong, as my counter-example shows.

Counter-example 2:
P = {Q|-3}, Q = {8|R}, R = {S|-1}, S = {7|U}, U = {1|-3}.

Your method gives as result initial count = -1 and move value = 2
My method as well as thermography give as result initial count = 0 and move value = 3

What criteria do you use to claim that the correct result is yours.
I understand that you found a 5-move sequence but this is wrong : in fact you have a 2-move sequence followed later by a 3-move sequence. You keep saying that your method is the best one, that I should use it (or maybe the thermography) and mine is surely wrong. Why instead you do not consider that other interesting ideas may exist? Why do you exclude that simpler methods may exist?
BTW, without knowing your method, I guess my method is nearer from yours than from thermography because I use also hypothesis. I use also reversal and I do not really understand why you decided to avoid the use of such powerful tool.

[RobertJasiek]

Uh, no, I do not claim that the correct result is mine but I bow to thermography. My result is just "consistent within my method for my definitions of values".

My method is the best one for practical application if faster / simpler methods are inapplicable. E.g., for simple local endgames, making a hypothesis works but is overkill; a simple gote / sente comparison will do.

Your method is not plainly wrong but you advertise using a (somewhat) rich environment without actually ever using this fact. You must prove your point by using your assumptions. In particular, you must explain why a "sufficiently" rich environment is possible to use instead of a rich, arbitrarily dense environment of thermography.

I do not exclude that simpler ideas exist:
- simple local endgame evaluation
- method of comparing the opponent's branches
- my pragmatic method of comparing counts
- my pragmatic method of comparing move values
- CGT reversal (simpler in the sense of simplifying the tree) by difference game

Whether your alleged method is simpler I do not know. First perform it well, then we can judge better. However, how can some (rather) rich environment ever be simpler than no environment?

I prefer to avoid the more powerful rich environment because play in them by means of proving theorems or constructing thermographs algebraically or graphically is hard. Each time, I have to look up how it works. A have done it for a few examples and proofs of theorems but just because I can do it with great effort does not mean that I would like it or even pretend simplicity of execution. In the sample for [22], you see a bit of my related "exercises", for which I have tried to keep the theory of thermography as simple as possible but it remains complicated with linear algebra at each calculation step.

[Gérard TAILLE]

Your method is not plainly wrong but you advertise using a (somewhat) rich environment without actually ever using this fact. You must prove your point by using your assumptions. In particular, you must explain why a "sufficiently" rich environment is possible to use instead of a rich, arbitrarily dense environment of thermography.

I see here some misunderstanding. OC it is not surprising because I am still in the building process of my method.
First of all I believe that my “sufficiently rich environment” is only another formal definition of the rich, arbitrarily dense environment. Anybody can see what a rich environment is and everybody can build a rich environment as rich as she wants. But, for a non-mathematician, going to the limit to reach an arbitrarily dense environment is not that obvious. Basically there are no difference.
It is true that my "move value" definition uses explicitly a “sufficiently rich environment”, but I also said that the “move value” itself depends only on the subtree of the position.
IOW I can build my “move value” using a calculation ignoring the environment and then I have only to verify that the result fulfilled my definition of “move value”.
BTW I never found a position whose miai value (calculated by thermography) does not fulfilled my famous property 1. Maybe it is obvious because I said my “sufficiently rich environment” is basically identical to the rich, arbitrarily dense environment of thermography. That means that I can keep my definition of move value as it stands (with “sufficiently rich environment”) and choose thermography as calculation method, waiting confidently for a counter example.
Why then I work on another method? Because thermography cannot be used in practice. The reason for that is that thermography is very ambitious, being able to give not only the “miai value” and the count but also information at whatever temperature.
In order to build a far simpler method my goal is only to calculate the move value (= miai value) and the count, and nothing else.

[RobertJasiek]

" I can build my “move value” using a calculation ignoring the environment and then I have only to verify that the result fulfilled my definition of “move value”."

Good aim. Please work out your method, especially how to verify using the rich environment!

"In order to build a far simpler method my goal is only to calculate the move value (= miai value) and the count, and nothing else."

I see, but calculating gains may also be useful.

[Gérard TAILLE]

" I can build my “move value” using a calculation ignoring the environment and then I have only to verify that the result fulfilled my definition of “move value”."

Good aim. Please work out your method, especially how to verify using the rich environment!

"In order to build a far simpler method my goal is only to calculate the move value (= miai value) and the count, and nothing else."

I see, but calculating gains may also be useful.

You may also wonder why I defined a move value (see viewtopic.php?p=277735#p277735) instead of using the already known miai value. Here again it's a matter of simplicity for a go player.
I expect that a rich environment at a given temperature is easy to undertand, and I expect that to have to play locally under a certain temperature called "move value" is also easy to understand. As a consequence a go player may easily imagine that, facing a position P1 and a position P2, if the move value of P1 is greater than the move value of P2, then it will quite often happen that a move in P1 will appear before a move in P2 to reach the best score.
In the other hand, explaining what a "miai value" is, needs far more than few lines and in addition the goal looks not that obvious for a go player (how a non theorician go player can really understand what the base of a mast means?)

[Gérard TAILLE]

What is the "move value" of a simple ko position A

Code: Select all

               A                       A
              / \                     / \
             /   \                   /   \
            B     C                 B     c
           /                       /
          /                       /
         D                       d

let's call c and d the counts of leaves C and D.
How to justify that the "move value" va of position A is equal to the value m = (d - c) / 3 ?

1) let's take first a temperature t > m
Let's take a rich environment Rich(t, N)
In this environment the score of the game if black decides to gain the ko is
score1 = d - g(Nt/N) - g((N-1)t/N) + g((N-2)t/N) - ...
while the score of the game when black plays in the environment and white finishes the ko is
score 2 = c + g(Nt/N) + g((N-1)t/N) - g((N-2)t/N) - ...
=> score 2 - score1 = c - d + 2(Nt/N) + 2((N-1)t/N) - 2((N-2)t/N) + …
=> score 2 - score1 = c - d + 2t + 2t/N ((N-1) - (N-2) + (N-3) – (N-4) …)
If N is odd then ((N-1) - (N-2) + (N-3) – (N-4) …) = (N-1)/2
And if N is even then ((N-1) - (N-2) + (N-3) – (N-4) …) = N/2
in any case we have ((N-1) - (N-2) + (N-3) – (N-4) …) ≤ N/2
=> score 2 - score1 ≤ c - d +2t + 2t/N (N/2) = c - d + 3t < c – d + 3m = 0
Eventually t > m => score2 > score1 => black plays in the environment to reach the best score.
2) With the same kind of reasoning you can also prove that
t < m => black must play in the ko to reach the best score
3) and 4) same result with white to play
Finally the “move value” of position A is va = m = (d - c) / 3
OC such proof is quite boring but I am very convinced it works.

[RobertJasiek]

Currently, I lack time to check your proof draft carefully but I get your badic construction. One thing I notice immediately that I do not understand.

You presume t>m to imply c - d + 3t < c – d + 3m. How?

[Gérard TAILLE]

Currently, I lack time to check your proof draft carefully but I get your badic construction. One thing I notice immediately that I do not understand.

You presume t>m to imply c - d + 3t < c – d + 3m. How?

Oops you are right Robert. Here is a correction:
Let's take a rich environment Rich(t, N)
In this environment the score of the game if black decides to gain the ko is
score1 = d - g(Nt/N) - g((N-1)t/N) + g((N-2)t/N) - ...
while the score of the game when black plays in the environment and white finishes the ko is
score 2 = c + g(Nt/N) + g((N-1)t/N) - g((N-2)t/N) - ...
=> score 2 - score1 = c - d + 2(Nt/N) + 2((N-1)t/N) - 2((N-2)t/N) + …
=> score 2 - score1 = c - d + 2t + 2t/N ((N-1) - (N-2) + (N-3) – (N-4) …)
If N is odd then ((N-1) - (N-2) + (N-3) – (N-4) …) = (N-1)/2
And if N is even then ((N-1) - (N-2) + (N-3) – (N-4) …) = N/2
in any case we have ((N-1) - (N-2) + (N-3) – (N-4) …) >= (N-1)/2
=> score 2 - score1 ≥ c - d + 2t + (2t/N)(N-1)/2) = c - d + 2t + t(N-1)/N
=> score 2 - score1 ≥ c - d + 3t – t/N
=> score 2 - score1 ≥ 3t – 3m – t/N
Choosing N1 > t / (3t -3m)
N > N1 => score 2 - score1 > 0
That way we can clearly see the consquence of being able to choose an environment as rich as you want.
Thank you Robert for your help.

[Gérard TAILLE]

Following the link given by jlt in the post viewtopic.php?p=277423#p277423 you can find the following position

$$
$$ | . . . , . . . . . , . .
$$ | . . . . . . . . . . . .
$$ | . . X . . . . . . . . .
$$ | . . . . O . . . . . . .
$$ | . . . X . . . . . . . .
$$ | . . . X O O . . . O . .
$$ | . . X , X O . . . , . .
$$ | . . . . X O . . . O . .
$$ | . . . X O a . . . . . . .
$$ | . . . . . . . . . . . .
$$ -------------------------

Click Here To Show Diagram Code

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

and the move value given is 20 points (deiri value).

My own analysis gives only 18 points and so I must be wrong somewhere.
What is your own result?

[Gérard TAILLE]

I have now made great progress in exploring my method to calculate move values and I am testing this method on various positions.
Taking the flashcards on the link gomagic-endgame-flashcards-lite-pack.zip I encounter some issues.
Here is one of them (number 24)

$$
$$ . . . . . . .|
$$ . . . . . . .|
$$ . . . . X . .|
$$ . . . . X . .|
$$ . . O O X . .|
$$ . . O X O . .|
$$ . . . . O . .|
$$ . . . O . . .|
$$ . . . . . . .|
$$ . . . . . . .|
$$

Click Here To Show Diagram Code

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

The site gives as answer 8 points gote (deiri value) where my result is 5 points in reverse sente <=> 10 points gote (deiri value).
What is your view?

[dfan]

Here are my calculations. No guarantee that they're correct, but they do match the flashcard.

$$B B+3 (the marked stones are Black's privilege)
$$ . . . . . . .|
$$ . . . . . . .|
$$ . . . . X T T|
$$ . . . . X T T|
$$ . . O O X 3 T|
$$ . . O X O 1 #|
$$ . . . . O 2 #|
$$ . . . O . @ @|
$$ . . . . . T T|
$$ . . . . . . .|
$$

Click Here To Show Diagram Code

[go]$$B B+3 (the marked stones are Black's privilege)
$$ . . . . . . .|
$$ . . . . . . .|
$$ . . . . X T T|
$$ . . . . X T T|
$$ . . O O X 3 T|
$$ . . O X O 1 #|
$$ . . . . O 2 #|
$$ . . . O . @ @|
$$ . . . . . T T|
$$ . . . . . . .|
$$[/go]

$$W W+5 (the marked stones are an even split)
$$ . . . . . . .|
$$ . . . . . . .|
$$ . . . . X T T|
$$ . . . . X 2 #|
$$ . . O O X 1 @|
$$ . . O X O 3 T|
$$ . . . . O T T|
$$ . . . O . T T|
$$ . . . . . T T|
$$ . . . . . . .|
$$

Click Here To Show Diagram Code

[go]$$W W+5 (the marked stones are an even split)
$$ . . . . . . .|
$$ . . . . . . .|
$$ . . . . X T T|
$$ . . . . X 2 #|
$$ . . O O X 1 @|
$$ . . O X O 3 T|
$$ . . . . O T T|
$$ . . . O . T T|
$$ . . . . . T T|
$$ . . . . . . .|
$$[/go]

That gives a total swing of 8 points in gote.