Life In 19x19

Posted: **Sat Jan 13, 2024 1:49 am**

I'm not sure if it qualifies as big endgame but I had an idea for a practical estimation for a common position. I'm not sure if it is a good estimate, maybe someone has suggestions as to how to test if it is a good estimate and how to improve it?

The position is this first line hane. Sometimes this is a double sente in practice. However, a double sente means one player draws the short straw when 4 points go to the other player.

Click Here To Show Diagram Code: [go]$$B $$ ------------------+ $$ . . . . . . . . .| $$ . . . . . . . 4 .| $$ . . O . . . . . 2| $$ . . . . O O O O 1| $$ . . . X X X X X 3| $$ . . . . . . . . .| $$ . . . . . . . . .| $$ . . . . . . . . .| $$ . . . . . . X . .| $$ . . . . . , . . .|[/go]

Besides treating this as a double sente we try treating it as a reverse sente, which would work when we have an opinion on whose reverse sente it is, and finally we could treat it like gote. I don't think there is much to say about the reverse sente case but unless it is clear that it is sente for one side it doesn't help. The gote case feels harder.

Here is were I had the dangerous idea to draw the number from my hat. First I assigned a round number to the following positions.

Click Here To Show Diagram Code: [go]$$W -10 $$ ------------------+ $$ . . . . . . . . .| $$ . . . . . . . . .| $$ . . O . . . . . .| $$ . . . . O O O O O| $$ . . . X X X X X O| $$ . . . . . . . 1 X| $$ . . . . . . 2 . .| $$ . . . . . . . . .| $$ . . . . . . X . .| $$ . . . . . , . . .|[/go]

Click Here To Show Diagram Code: [go]$$B +10 $$ ------------------+ $$ . . . . . . . . .| $$ . . . . . . 4 . 3| $$ . . O . . . 2 1 O| $$ . . . . O O O O X| $$ . . . X X X X X X| $$ . . . . . . . . .| $$ . . . . . . . . .| $$ . . . . . . . . .| $$ . . . . . . X . .| $$ . . . . . , . . .|[/go]

Then I wrote
G := { { 10 + A | 10 } | { -10 | -10 + B } }
and declared that the temperature is
T = 10 + (A - B) / 4

A and B could be interpreted in terms of the reduction of the territory in the diagrams is when the second player doesn't defend. That is if the reduction from a second black, or white, move is 15 points, or -15, then A, or B, would be 5, or -5. Of course one could use a better estimate than 10 and there is not particular reason that it has to be symmetric. Btw 10 is a 3x3 square and a prisoner, something that is easy to see on the board.

How about it? Is this a useful approximation rule? Maybe for the less serious? What is the error in the estimate?

Posted: **Sun Jan 14, 2024 5:46 am**

kvasir wrote:I'm not sure if it qualifies as big endgame but I had an idea for a practical estimation for a common position. I'm not sure if it is a good estimate, maybe someone has suggestions as to how to test if it is a good estimate and how to improve it?

Environmental Go( https://senseis.xmp.net/?EnvironmentalGo looks like it would be good and fun way to test it. Is there any way to use AI programmes with Environmental Go, in place of a couple of professional players?

Posted: **Sun Jan 14, 2024 3:19 pm**

I think I meant to write
G := { { 10 + A | 2 } | { -2 | -10 + B } }
which would (if I didn't make more mistakes) lead to the approximation rule
T ~= 6 + (A – B) / 4

I did some exploration with KataGo which led to the following form for the game
G := { { 17 | 2 } | { -2 | -15 } }

Note that I have rounded the numbers in an attempt at avoiding the illusion of accuracy.

I identified two significant elements that affect how the position is played out:

What other moves are available. There must be some large moves available to justify exploring the gote outcomes. However, there are also two possible ko that KataGo will try to exploit.
The safety of the groups. The continuations that I showed are from professional games that I could find similar situations in and naturally the groups in those games were different than the artificial situation that I presented to KataGo. That is the follow ups are slightly different.

As a consequence it is difficult to isolate the evaluation from detail that only apples in the exact position used for the exploration. Still I tried to do just that.

What I did is that I started from a scaffold game
G_scaffold := { { X | 2 } | { -2 | Y } }
and estimated X - 2 and Y - 2 by offering KataGo to capture dango in lieu of playing in the position. This is, arguably, appropriate since we know the values for the cases when the position is defended in gote and it is straight forward to observe when KataGo prefers to defended in gote and when it will capture the dango.

By this means I arrived at
G = { { 17 | 2 } | { -2 | -15 } }

Based on this game T = 9. This is not far off from the point loss of playing pass, which was 9.8. The pass-estimate depends very much on the available moves.

Now for the approximation rule that was suggested.

G := { { 10 + A | 2 } | { -2 | -10 + B } }
T ~= 6 + (A – B) / 4

Here A = 7 and B = -5 so
T ~= = 9

That is of course the same estimate as above but possibly it is easier to estimate A and B visually then it is to estimate the move value directly?

Posted: **Sun Jan 14, 2024 4:22 pm**

How might one get an estimate for A and B without too much trouble?

For A

1. Visualize a correct continuation.

Pick the most suitable continuation, if you make simplification just make sure it is close in value to whatever would be the 100% move. Here I assume the ko variation isn't that much different and use the easier variation.

Click Here To Show Diagram Code: [go]$$B Correct sequence $$ ------------------+ $$ . . . . . . . . .| $$ . . . . . . . . 3| $$ . . O . . . 2 1 O| $$ . . . . O O O O X| $$ . . . X X X X X X| $$ . . . . . . . . .| $$ . . . . . . . . .| $$ . . . . . . . . .| $$ . . . . . . X . .| $$ . . . . . , . . .| $$ . . . . . . . . .|[/go]

2. Remember that sente follow ups are treated as if they can always be played.

Click Here To Show Diagram Code: [go]$$B Sente follow up $$ ------------------+ $$ . . . . 4 3 5 . .| $$ . . . . 6 2 1 . X| $$ . . O . . . O X .| $$ . . . . O O O O X| $$ . . . X X X X X X| $$ . . . . . . . . .| $$ . . . . . . . . .| $$ . . . . . . . . .| $$ . . . . . . X . .| $$ . . . . . , . . .| $$ . . . . . . . . .|[/go]

3. Visualize the excess points reduced (or made) in the variation beyond the 10 points.

Click Here To Show Diagram Code: [go]$$B $$ ------------------+ $$ . . . . P Z X M M| $$ . . . . P P X M X| $$ . . O . . . O X .| $$ . . . . O O O O X| $$ . . . X X X X X X| $$ . . . . . . . . .| $$ . . . . . . . . .| $$ . . . . . . . . .| $$ . . . . . . X . .| $$ . . . . . , . . .| $$ . . . . . . . . .|[/go]

I get A = 7

For B

1. Visualize a correct continuation.

Again picking the follow up that I find easiest to work with.

Click Here To Show Diagram Code: [go]$$W $$ ------------------+ $$ . . . . . . . . .| $$ . . . . . . . . .| $$ . . O . . . . . .| $$ . . . . O O O O O| $$ . . . X X X X X O| $$ . . . . . . . 1 X| $$ . . . . . . . . 3| $$ . . . . . . . 2 .| $$ . . . . . . X . .| $$ . . . . . , . . .| $$ . . . . . . . . .|[/go]

2. Remember that sente follow ups are treated as if they can always be played.

Click Here To Show Diagram Code: [go]$$W B4 missing $$ ------------------+ $$ . . . . . . . . .| $$ . . . . . . . . .| $$ . . O . . . . . .| $$ . . . . O O O O O| $$ . . . X X X X X O| $$ . . . . . . . O .| $$ . . . . 4 3 1 . O| $$ . . . . . 2 5 X .| $$ . . . . . . X 6 .| $$ . . . . . , . . .| $$ . . . . . . . . .|[/go]

3. Visualize the excess points reduced (or made) in the variation beyond the 10 points.

Click Here To Show Diagram Code: [go]$$B $$ ------------------+ $$ . . . . . . . . .| $$ . . . . . . . . .| $$ . . O . . . . . .| $$ . . . . O O O O O| $$ . . . X X X X X O| $$ . . . . . M . O .| $$ . . . . Z P O . O| $$ . . . . . Z O X .| $$ . . . . . . X Z .| $$ . . . . . , . . .| $$ . . . . . . . . .|[/go]

I get B = -5

Conclusion

Maybe you realized that the value 10 is arbitrary and one could use other numbers depending on the context, if one adjusted the formula correctly. Here 10 was a nice number since it forms an easy to visualize square on the board.

Another thing to realize is that excess reduction is divided by 4 in the approximation rule and for every 1 point of error in the final you'd need to be off by 4 points in the total for the visualized estimate. The error is likely reduced.

If we assume the error in estimating any position is a Guassian with variance s*s, then the error in the approximation formula should be a Guassian with variance 0.5 * s*s. Maybe this is not the right way to analyze the error, I'm not sure, but note that in a more brute force approach we would estimate many more positions and follow ups. That would be error prone and any error would cascade, but in this case we seem to get away with only to evaluations (even if that may also introduce error).

Now I could have estimated A = 8 and B = -6 based on other possible follow ups. My original choice was more valid in the whole board position that I explored with KataGo but it is tricky and other follow ups may be more likely when things are different. The approximation rule would give T ~= 9.5 and that is not far off for the other estimate.

Posted: **Thu Jan 18, 2024 5:10 am**

Your central idea is division by 4, which applies to either starting player making 2 excess plays resulting in the tally 4. So it is treated as a double-gote for both.

I have described this idea on 2019-04-20 in [29], see https://www.lifein19x19.com/viewtopic.p ... 45#p143245

You want to apply to shapes open on the outside, where some assumptions must be made on how boundaries might settle.

Posted: **Sat Jan 20, 2024 3:37 am**

Your thread title has a very much bigger scope than your special study. Other major approximations for the big / early endgame are as follows and described and solved in general for arbitrary comparatively simple local shapes in arbitrary idealised global environments in [22], [23]:

- My theorem for Bill Spight's idea of the T/2 estimate of moving first in the environment at temperature T.
- Elwyn Berlekamp's theorems with the large error T/2 for the strategy sentestrat.
- My theorems with the small error 2D for the timing of gote and the drop D of move values.
- My theorem of delaying sente at high temperature (that is, larger than the follow-up move value).
- My theorems for big local endgames with one follow-up during the early endgame at low temperature.
- My and partly Bill Spight's theorems for big local endgames with gote and sente options.
- My theorems for big local endgames (of the types a) with one follow-up or b) gote with two follow-ups) during the late endgame are solutions of correct play. However, they can be applied as approximations if the environment is not simple.

Posted: **Sat Feb 03, 2024 8:57 am**

I think it is a good idea to see how to approximate a big endgame and I guess each player has it's own method.
Let't try to understand yours.

I begin with the position after black hane.

Click Here To Show Diagram Code: [go]$$B +10 $$ ------------------+ $$ . . . . . . . . .| $$ . . . . . . 4 . 3| $$ . . O . . . 2 1 O| $$ . . . . O O O O X| $$ . . . X X X X X X| $$ . . . . . . . . .| $$ . . . . . . . . .| $$ . . . . . . . . .| $$ . . . . . . X . .| $$ . . . . . , . . .|[/go]

You claim here that the count is +10 and I agree with you by assuming the following sequence:

Click Here To Show Diagram Code: [go]$$W $$ ------------------+ $$ . . . . . . . . .| $$ . . . . . . O 1 X| $$ . . O . . . O X 2| $$ . . . . O O O O X| $$ . . . X X X X X X| $$ . . . . . . . . .| $$ . . . . . . . . .| $$ . . . . . . . . .| $$ . . . . . . X . .| $$ . . . . . , . . .|[/go]

First of all I assume white will exchange in sente

for

Click Here To Show Diagram Code: [go]$$B $$ ------------------+ $$ . . . . . . 4 3 .| $$ . . . . . . O O X| $$ . . O . . . O X X| $$ . . . . O O O O X| $$ . . . X X X X X X| $$ . . . . . . . . .| $$ . . . . . . . . .| $$ . . . . . . . . .| $$ . . . . . . X . .| $$ . . . . . , . . .|[/go]

then I assume black will exchange in sente

for

Click Here To Show Diagram Code: [go]$$B $$ ------------------+ $$ . . . . . . O X 5| $$ . . . . . . O O X| $$ . . O . . . O X X| $$ . . . . O O O O X| $$ . . . X X X X X X| $$ . . . . . . . . .| $$ . . . . . . . . .| $$ . . . . . . . . .| $$ . . . . . . X . .| $$ . . . . . , . . .|[/go]

and finally I agree to simplify the count by assuming black will connect the ko with

Taking into account a white prisonner it is clear that the count of the position is +10.

Now what about your calculation of A?

You assumed the following sequence.

Click Here To Show Diagram Code: [go]$$B Sente follow up $$ ------------------+ $$ . . . . 4 3 5 . .| $$ . . . . 6 2 1 . X| $$ . . O . . . O X .| $$ . . . . O O O O X| $$ . . . X X X X X X| $$ . . . . . . . . .| $$ . . . . . . . . .| $$ . . . . . . . . .| $$ . . . . . . X . .| $$ . . . . . , . . .| $$ . . . . . . . . .|[/go]

The count of the position is for me +18 (13 points of territory for white + 4 points of territory for black + 1 white prisonner).

For me A = 8 (instead of 7).

BTW I found B = 7 (instead of 5), showing our calculations are quite different. Could you clarify your calculations?

Considering now the basis of your post I do not understand how your formula T = 10 + (A - B) / 4 can really work but let's begin by understanding first the count calculation.

Posted: **Sat Feb 03, 2024 11:05 am**

Gerard, corrected my formula above.

kvasir wrote: I think I meant to write
G := { { 10 + A | 2 } | { -2 | -10 + B } }
which would (if I didn't make more mistakes) lead to the approximation rule
T ~= 6 + (A – B) / 4

In that case you have A = 8 and B = -7, for

Code: Select all

T ~= 6 + (A – B) / 4
   = 6 + (8 – (-7)) / 4
   = 9.75

Do you see how T ~= 9.75 is not far from say T ~= 9 when talking about the big endgame? It is possible that 9.75 is more correct, my pass-move exploration with KataGo suggested an evaluation T ~= 9.8.

The approximation formula is based on realizing that there are two problem that need to be solved implicitly to have a good approach in the big endgame. First that the temperature for large endgames is usually highly dependent on the exact position. Second is that we make more mistakes the more variations we look at.

As to the first. The big endgame is especially wrought with pitfalls. The timing of a move is highly dependent on the position in the particular game. A move may be played earlier if some follow ups are likely to be sente and it will be played later if the follow ups are are likely to be gote, but that isn't complete, it also matters if there will be other moves of similar value and if the number of such moves is odd or even. It can be very intricate.

As to the second. There are a number of sources of errors. Simply counting incorrectly or incorrectly copying a number is a simple mistakes that happens all the time. A more complex source of errors is that you may look at the wrong moves, they may not have been the best moves. The source of errors that always baffles myself the most is that you need to know if something will be sente or gote later in the game, that is not always easy.

I don't think there is any one solution to how avoid these errors but my experience tells me that the more variations and the more counting you do the more likely it is that you make an error.

In that order I proposed to approximate this apparent double sente positions in a way that simplifies the task and avoids to look at many variations.

The more generic method is that we start with:
a, the size of sente reduction for black
big_A, an easy to visualize area that black targets for reduction
curvv_A, a simple estimate of how much black reduces if white tenuki
b, the size of sente reduction for white
big_B, an easy to visualize area that white targets for reduction
curvv_B, a simple estimate of how much white reduces if black tenuki

This would give a game scaffold like this

G := { { curvv_A | a } | { b | curvv_B } }

Which I rewrote by introducing A = curvy_A - big_A and B = curvy_B - big_B as

G = { { A + big_A | a } | { b | A + big_B } }

now it follows that the temperature for this game is

T = (big_A + a - (b + big_B)) / 4 + (A - B) / 4

Now I suggest that we can look up (big_A + a - (b + big_B)) in a table, at least we can come up with a value quickly.

In my example we had
a = 2
big_A = 10
b = -2
big_B = -10

and the formula simplifies to
T = 6 + (A - B) / 4

Now I prefer to write ~= instead of = to indicate that this is not indented to be exact
T ~= 6 + (A - B) / 4
and if you prefer to work with positive numbers (which I do think is more natural in this example) it should be
T ~= 6 + (A + B) / 4

Maybe this goes some way toward answering what the method is?

The benefit here is that you end up only estimating curvy_A and curvey_B. In my experience it is much more likely that errors are made when many things need to be estimated carefully. That brings me to the second benefit, the difficult estimation task produces values that are divided by 4, if a small error is acceptable (which I argue it is when talking about such large endgame moves) then this makes it easier to control the error.

I'm sure I made many errors in the posts above, for example I wrote "apples" when I meant "applies"

Maybe I should proof read and correct.

Posted: **Sat Feb 03, 2024 11:39 am**

kvasir wrote: I don't think there is any one solution to how avoid these errors but my experience tells me that the more variations and the more counting you do the more likely it is that you make an error. [...] Maybe I should proof read and correct.

A solution for reducing or avoiding errors depends on the person.

I have written thousands of published pages on endgame calculations and mathematical theory, AFAIK, without one mistake. Having the mindset of a mathematician is very useful for finding and correcting all mistakes. It is the duty of a mathematician to write correct contents only. During drafts, I make mistakes in calculations, statements or proofs. All such mistakes I eliminate by proofreading often enough, which I may assess when proofreading about thrice in succession without finding any further mistake.

In forum messages, one hardly works so rigorously. However, you get the idea: a principle ability of finding all one's mistakes is required; work the most diligently.

Posted: **Sat Feb 03, 2024 2:27 pm**

kvasir wrote: I think I meant to write
G := { { 10 + A | 2 } | { -2 | -10 + B } }
which would (if I didn't make more mistakes) lead to the approximation rule
T ~= 6 + (A – B) / 4
...
The source of errors that always baffles myself the most is that you need to know if something will be sente or gote later in the game, that is not always easy.

Let's take only the initial black branch.
You claim the tree corresponds to { 10 + A | 2 } and, logically the black branch count is then 6 + A/4.
My view is that { 10 + A | 2 } is true ONLY if A <= 8. if A > 8 then the tree can be reduced to only {10}.
As a consequence the count of black branch is min(6 + A/4, 10) instead of 6 + A/4.
You mentionned yourself that sente and gote relationship need to be analysed. Here it is the case.

Click Here To Show Diagram Code: [go]$$B +10 $$ ------------------+ $$ . . . . . . . . .| $$ . . . . . . . . 1| $$ . . O . . . O X O| $$ . . . . O O O O X| $$ . . . X X X X X X| $$ . . . . . . . . .| $$ . . . . . . . . .| $$ . . . . . . . . .| $$ . . . . . . X . .| $$ . . . . . , . . .|[/go]

if

is sente the count of black branch is 10 and if

is gote then the value of the branch is 6+A/2.
if A > 8 then

is sente, if A < 8 then

is gote and if A = 8 the status is ambiguous.
I am sure you can verify that your formula 6 + A/4 is not correct if A = 9.

Posted: **Sat Feb 03, 2024 5:38 pm**

Gérard TAILLE wrote:if is sente the count of black branch is 10 and if is gote then the value of the branch is 6+A/2.
if A > 8 then is sente, if A < 8 then is gote and if A = 8 the status is ambiguous.
I am sure you can verify that your formula 6 + A/4 is not correct if A = 9.

OK, I see what you mean.

The temperature of the { reverse-sente | gote } case
G1 = { 10 | { -2 | -10 + B } }
and the { gote | gote } case
G2 = { { 10 + A | 2 } | { -2 | -10 + B } }
is the same if A == 8.

So the formula appears to need adjustment if we have reverse sente continuations. I don't think I addressed that sufficiently in the previous messages, I don't remember what I wrote exactly, but basically the denominators would have to change if you pick a sente continuation when the defender plays away.

On first impression it appears to me that depending on the case these formulas would apply.

G = { gote | gote }
T = 6 + (A - B) / 4

G = { gote | reverse-sente }
T = 8 + A / 4 - B / 2

G = { reverse-sente | gote }
T = 8 + A / 2 - B / 4

G = { reverse-sente | reverse-sente}
T = 10 + (A - B) / 2

Now if we are choosing between
T = 6 + (A - B) / 4
and
T' = 8 + A' / 2 - B / 4
then black should choose T if T > T' and T' if T < T'.

If we let A = 8 and A' = 0 then T = T'. This also occurs if A = 4 and A' = -2, and for many other values and it is possible to pick an incorrect case.

Posted: **Sun Feb 04, 2024 4:51 am**

kvasir wrote:
Gérard TAILLE wrote:if is sente the count of black branch is 10 and if is gote then the value of the branch is 6+A/2.
if A > 8 then is sente, if A < 8 then is gote and if A = 8 the status is ambiguous.
I am sure you can verify that your formula 6 + A/4 is not correct if A = 9.
OK, I see what you mean.

The temperature of the { reverse-sente | gote } case
G1 = { 10 | { -2 | -10 + B } }
and the { gote | gote } case
G2 = { { 10 + A | 2 } | { -2 | -10 + B } }
is the same if A == 8.

So the formula appears to need adjustment if we have reverse sente continuations. I don't think I addressed that sufficiently in the previous messages, I don't remember what I wrote exactly, but basically the denominators would have to change if you pick a sente continuation when the defender plays away.

On first impression it appears to me that depending on the case these formulas would apply.

G = { gote | gote }
T = 6 + (A - B) / 4

G = { gote | reverse-sente }
T = 8 + A / 4 - B / 2

G = { reverse-sente | gote }
T = 8 + A / 2 - B / 4

G = { reverse-sente | reverse-sente}
T = 10 + (A - B) / 2

Now if we are choosing between
T = 6 + (A - B) / 4
and
T' = 8 + A' / 2 - B / 4
then black should choose T if T > T' and T' if T < T'.

If we let A = 8 and A' = 0 then T = T'. This also occurs if A = 4 and A' = -2, and for many other values and it is possible to pick an incorrect case.

That sound better but for me far too complicate in practice. How do you find quickly the values 8, 10 or 6 appearing in the T formulas? In addition how do you see quickly a gote or a sente or a reverse sente situation? My own method is completly different. I do not claim it is better than yours, but in such situation it works particularly well.

For the time being I showed you some mistakes in the branch black moving first . It is time to show you the branch white to move with a hane.

Click Here To Show Diagram Code: [go]$$W $$ ------------------+ $$ . . . . . . . . .| $$ . . . . . . . . .| $$ . . O . . . . . .| $$ . . . . O O O O O| $$ . . . X X X X X O| $$ . . . . . . . . X| $$ . . . . . . . . .| $$ . . . . . . . . .| $$ . . . . . . X . .| $$ . . . . . , . . .| $$ . . . . . . . . .|[/go]

Using the same same approach for this branch you visualize the white sente continuation:

Click Here To Show Diagram Code: [go]$$W $$ ------------------+ $$ . . . . . . . . .| $$ . . . . . . . . .| $$ . . O . . . . . .| $$ . . . . O O O O O| $$ . . . X X X X X O| $$ . . . . . . 4 1 X| $$ . . . . . . 6 5 3| $$ . . . . . . . 2 .| $$ . . . . . . X . .| $$ . . . . . , . . .| $$ . . . . . . . . .|[/go]

and you can easily assume to continue by

Click Here To Show Diagram Code: [go]$$B $$ ------------------+ $$ . . . . . . . . .| $$ . . . . . . . . .| $$ . . O . . . . . .| $$ . . . . O O O O O| $$ . . . X X X X X O| $$ . . . . . . X O 2| $$ . . . . . . X O O| $$ . . . . . . . X 1| $$ . . . . . . X . .| $$ . . . . . , . . .| $$ . . . . . . . . .|[/go]

you can see that the result is a count = -9 (instead of -10 in your calculation).

If now black prefers to keep sente then black must avoid playing

and the resulting sequence is the one you showed yourself:

Click Here To Show Diagram Code: [go]$$W B4 missing $$ ------------------+ $$ . . . . . . . . .| $$ . . . . . . . . .| $$ . . O . . . . . .| $$ . . . . O O O O O| $$ . . . X X X X X O| $$ . . . . . . . O .| $$ . . . . 4 3 1 . O| $$ . . . . . 2 5 X .| $$ . . . . . . X 6 .| $$ . . . . . , . . .| $$ . . . . . . . . .|[/go]

followed by

Click Here To Show Diagram Code: [go]$$B $$ ------------------+ $$ . . . . . . . . .| $$ . . . . . . . . .| $$ . . O . . . . . .| $$ . . . . O O O O O| $$ . . . X X X X X O| $$ . . . . . 3 2 O 4| $$ . . . . X O O 1 O| $$ . . . . . X O X .| $$ . . . . . . X X .| $$ . . . . . , . . .| $$ . . . . . . . . .|[/go]

and finally

Click Here To Show Diagram Code: [go]$$W $$ ------------------+ $$ . . . . . . . . .| $$ . . . . . . . . .| $$ . . O . . . . . .| $$ . . . . O O O O O| $$ . . . X X X X X O| $$ . . . . . X O O O| $$ . . . . X O O X O| $$ . . . . . X O X 1| $$ . . . . . . X X 2| $$ . . . . . , . . .| $$ . . . . . . . . .|[/go]

and the count is now -16

taking your formulas
G = { gote | gote }
T = 6 + (A - B) / 4
and according to the sequences above I think the correct formula is:
G = { gote | gote }
T = 5.75 + (A - B) / 4 with A = 8 and B = -7 => T = 9.5

Do you agree?

Posted: **Sun Feb 04, 2024 10:59 am**

kvasir wrote: On first impression it appears to me that depending on the case these formulas would apply.

G = { gote | gote }
T = 6 + (A - B) / 4

G = { gote | reverse-sente }
T = 8 + A / 4 - B / 2

In my previous post I looked only on your first formula :
G = { gote | gote }
T = 6 + (A - B) / 4

Now I looked also to your other formulas which seem to me not correct.
Instead of
G = { gote | reverse-sente }
T = 8 + A / 4 - B / 2

my view is
G = { gote | reverse-sente }
T = 8 + A / 4
but maybe I have not a good understanding of what you mean by G = { gote | reverse-sente }.
Could you explain a little more how you reach your formula?

Posted: **Sun Feb 04, 2024 11:02 am**

Gérard TAILLE wrote:That sound better but for me far too complicate in practice. How do you find quickly the values 8, 10 or 6 appearing in the T formulas? [...]

Like I said you would have the cases that are useful prepared in advance if you think that is useful for estimating some big endgame moves.

Gérard TAILLE wrote:[...] In addition how do you see quickly a gote or a sente or a reverse sente situation? My own method is completly different. I do not claim it is better than yours, but in such situation it works particularly well.

That is what Go players are skilled at doing. The idea here is that Go players can work with a small number of variation and you can memorize one or more cases that are useful for simplifying the approximation. It is a mnemonic, you memorize a few cases and then you can apply the approximation rule for those cases.

Gérard TAILLE wrote:you can see that the result is a count = -9 (instead of -10 in your calculation).

No. I did not count -10, I picked that number for the reason that it is easy to remember and easy to visualize.

Gérard TAILLE wrote:taking your formulas
G = { gote | gote }
T = 6 + (A - B) / 4
and according to the sequences above I think the correct formula is:
G = { gote | gote }
T = 5.75 + (A - B) / 4 with A = 8 and B = -7 => T = 9.5

Do you agree?

I don't think I do. I didn't pick the number 10 for the reason that you are now assuming. The values for A and B also depend on the exact position and you your ability to estimate. Of course you could pick any number but I picked the numbers for the reason that I wanted to make it easy to visualize, not because it is derived from any specific variation.

For example, one could write
G = { { 100 + A | 2 } | { -2 | -100 – B } }
T ~= 51 + (A – B) / 4

and I don't think that would be useful. Ten unlike one hundred can easily be visualized, that conclusion is based on introspection. You could use nine but that would not give whole numbers.

Further, there is no one correct value for temperature or the subgames that is always correct. This particular endgame does depend on things like if there is a possibility to escape into the center and on ko fights, but it appears to me that you are arguing which variations are the correct variations in general.

To turn back to the OP I wrote...

kvasir wrote: I'm not sure if it qualifies as big endgame but I had an idea for a practical estimation for a common position. I'm not sure if it is a good estimate, maybe someone has suggestions as to how to test if it is a good estimate and how to improve it?

...and (here the numbers need some correction)...

kvasir wrote:A and B could be interpreted in terms of the reduction of the territory in the diagrams is when the second player doesn't defend. That is if the reduction from a second black, or white, move is 15 points, or -15, then A, or B, would be 5, or -5. Of course one could use a better estimate than 10 and there is not particular reason that it has to be symmetric. Btw 10 is a 3x3 square and a prisoner, something that is easy to see on the board.

How about it? Is this a useful approximation rule? Maybe for the less serious? What is the error in the estimate?

I'm interesting what you think about this approach. Do you understand my idea and why it could be useful, should I try to present it more precisely? I don't know if or when I'd do that.

I'm also interested in your ideas about how to verify such rules, both based on comparing with exact estimates and based the kinds of errors that are possible.

Posted: **Sun Feb 04, 2024 11:26 am**

Gérard TAILLE wrote:
kvasir wrote: On first impression it appears to me that depending on the case these formulas would apply.

G = { gote | gote }
T = 6 + (A - B) / 4

G = { gote | reverse-sente }
T = 8 + A / 4 - B / 2
In my previous post I looked only on your first formula :
G = { gote | gote }
T = 6 + (A - B) / 4

Now I looked also to your other formulas which seem to me not correct.
Instead of
G = { gote | reverse-sente }
T = 8 + A / 4 - B / 2

my view is
G = { gote | reverse-sente }
T = 8 + A / 4
but maybe I have not a good understanding of what you mean by G = { gote | reverse-sente }.
Could you explain a little more how you reach your formula?

That was my initial form also, but my impression was that it should be T = 8 + A / 4 - B / 2, but let me try to check it more carefully.

The forms for gote and reverse-sente should be

gote = { L | R } , where L and R are some values, that is the temperature of this game is T = (L - R) / 2 and the value v = (L + R) / 2

reverse-sente = { { | L } | R } , if that makes sense, that is the temperature of this game would be T = L - R and the value v = L.

I'm sorry if I have misused the notation, it is not unlikely that I have.

Basically, for the G = { gote | reverse-sente } case I let

G_L := { 10 + A | 2 }
G_R := { -2 | B - 10 }
G := { G_L | G_R }

Now in terms of value and temperature, but not structure, we have
G = { { 10 + A | 2 } | B - 10 }
or alternatively
G = { 6 + A / 2 | B - 10 }
and the temperature would then be
T = 8 + A / 4 - B / 2

Is my reasoning understandable? Is it correct?

Life In 19x19

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