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 Post subject: Re: How evaluate double sente moves ?
Post #161 Posted: Sun Oct 25, 2020 5:20 am 
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Gérard TAILLE wrote:
Bill Spight wrote:
If you don't want to be vague, let t = 0. But anyway, heuristics can help guide reading.

If you want 2 ≤ t < 3, then replace {8|5} with 6½ and play G1. If you are not more specific, what can be said about best play?


OK Bill let's simplify the problem and formulate it in a different way.

Let's take G1 = {8|5||0} and G2 = {6|0}
and let's assume G3,G4,.. only of the forms {x|y} (I mean only simple gote) with g1 ≥ g2 ≥ g3 ≥ g4 ...
In addition let's suppose the game black to play G1 + G2 + G3 + G4 + ... is quite close and let's suppose that the god play allows black to win the game.
Question : what is the best way to proceed in order to be sure at 100% to choose as a first black move a winning move?


OK. Without loss of generality, let Gi = {gi|-gi}. Let gi' = gi - g(i+1) + g(i+2) - ....

I. g3 ≤ 1½
A. Black to play
1. G1 first: Score1 = 6½ - 0 + 1½ - g3' = 8 - g3'
2. G2 first: Score2 = 6 - 0 + g3' = 6 + g3'
Diff = Score1 - Score2 = 2 - 2g3'
-1 ≤ Diff ≤ 2
B. White to play
1. G1 first: Score2 = 0 + 6 - g3' = 6 - g3'
1. G2 first: Score1 = 0 + 6½ - 1½ + g3' = 5 + g3'
Diff = Score2 - Score1 = -1 + 2g3' ; Both players try to maximize Diff
-1 ≤ Diff ≤ 2
II. g4 ≤ 1½ ≤ g3 ≤ 3
A. Black to play
1. G1 first: Score1 = 6½ - 0 + g3 - 1½ + g4' = 5 + g3 + g4'
2. G2 first: Score2 = 6 - 0 + g3 - g4' = 6 + g3 + g4'
Diff = -1 + 2g4'
-1 ≤ Diff ≤ 2
B. White to play
1. G1 first: Score1 = 0 + 6 - g3 + g4' = 6 - g3 + g4'
2. G2 first: Score2 = 0 + 6½ - g3 + 1½ - g4' = 8 - g3 - g4'
Diff = 2 - 2g4'
-1 ≤ Diff ≤ 2
III. g5 ≤ 1½ ≤ g4 ≤ g3 ≤ 3
A. Black first
1. G1 first: Score1 = 6½ + g3 - g4 + 1½ - g5' = 8 + g3 - g4 - g5'
2. G2 first: Score2 = 6 + g3 - g4 + g5'
Diff = 2 - 2g5'
B. White first
1. G1 first: Score1 = 6 - g3 + g4 - g5'
2. G2 first: Score1 = 5 - g3 + g4 + g5'
Diff = -1 + 2g5'
IV. g6 ≤ 1½ ≤ g5 ≤ g4 ≤ g3 ≤ 3
A. Black first
Diff = -1 + 2g6'
B. White first
Diff = 2 - 2g6'

Interesting pattern. :)

When g(2n+1) ≤ 1½ ≤ g(2n), Black to play plays in G1 when 2 ≥ 2g(2n+1)', and White to play plays in G1 when 2g(2n+1)' ≥ 1.

When g(2n) ≤ 1½ ≤ g(2n-1), Black to play plays in G1 when 2g(2n)' ≥ 1, and White to play plays in G1 when 2 ≥ 2g(2n)'.

I expect that you already knew that. :) It is part of CGT, but not thermography.

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 Post subject: Re: How evaluate double sente moves ?
Post #162 Posted: Sun Oct 25, 2020 6:23 am 
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Bill Spight wrote:
Gérard TAILLE wrote:
Bill Spight wrote:
If you don't want to be vague, let t = 0. But anyway, heuristics can help guide reading.

If you want 2 ≤ t < 3, then replace {8|5} with 6½ and play G1. If you are not more specific, what can be said about best play?


OK Bill let's simplify the problem and formulate it in a different way.

Let's take G1 = {8|5||0} and G2 = {6|0}
and let's assume G3,G4,.. only of the forms {x|y} (I mean only simple gote) with g1 ≥ g2 ≥ g3 ≥ g4 ...
In addition let's suppose the game black to play G1 + G2 + G3 + G4 + ... is quite close and let's suppose that the god play allows black to win the game.
Question : what is the best way to proceed in order to be sure at 100% to choose as a first black move a winning move?


OK. Without loss of generality, let Gi = {gi|-gi}. Let gi' = gi - g(i+1) + g(i+2) - ....

I. g3 ≤ 1½
A. Black to play
1. G1 first: Score1 = 6½ - 0 + 1½ - g3' = 8 - g3'
2. G2 first: Score2 = 6 - 0 + g3' = 6 + g3'
Diff = Score1 - Score2 = 2 - 2g3'
-1 ≤ Diff ≤ 2
B. White to play
1. G1 first: Score2 = 0 + 6 - g3' = 6 - g3'
1. G2 first: Score1 = 0 + 6½ - 1½ + g3' = 5 + g3'
Diff = Score2 - Score1 = -1 + 2g3' ; Both players try to maximize Diff
-1 ≤ Diff ≤ 2
II. g4 ≤ 1½ ≤ g3 ≤ 3
A. Black to play
1. G1 first: Score1 = 6½ - 0 + g3 - 1½ + g4' = 5 + g3 + g4'
2. G2 first: Score2 = 6 - 0 + g3 - g4' = 6 + g3 + g4'
Diff = -1 + 2g4'
-1 ≤ Diff ≤ 2
B. White to play
1. G1 first: Score1 = 0 + 6 - g3 + g4' = 6 - g3 + g4'
2. G2 first: Score2 = 0 + 6½ - g3 + 1½ - g4' = 8 - g3 - g4'
Diff = 2 - 2g4'
-1 ≤ Diff ≤ 2
III. g5 ≤ 1½ ≤ g4 ≤ g3 ≤ 3
A. Black first
1. G1 first: Score1 = 6½ + g3 - g4 + 1½ - g5' = 8 + g3 - g4 - g5'
2. G2 first: Score2 = 6 + g3 - g4 + g5'
Diff = 2 - 2g5'
B. White first
1. G1 first: Score1 = 6 - g3 + g4 - g5'
2. G2 first: Score1 = 5 - g3 + g4 + g5'
Diff = -1 + 2g5'
IV. g6 ≤ 1½ ≤ g5 ≤ g4 ≤ g3 ≤ 3
A. Black first
Diff = -1 + 2g6'
B. White first
Diff = 2 - 2g6'

Interesting pattern. :)

When g(2n+1) ≤ 1½ ≤ g(2n), Black to play plays in G1 when 2 ≥ 2g(2n+1)', and White to play plays in G1 when 2g(2n+1)' ≥ 1.

When g(2n) ≤ 1½ ≤ g(2n-1), Black to play plays in G1 when 2g(2n)' ≥ 1, and White to play plays in G1 when 2 ≥ 2g(2n)'.

I expect that you already knew that. :) It is part of CGT, but not thermography.


Oops maybe you missed the point. The problem was not to find the best move but only to find a winning move!
My own result is far simplier:
if the score 6 + g3' allows me to win the game then my first play will be in G2 otherwise I will play in G1.
Could you verify if I am wrong?

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 Post subject: Re: How evaluate double sente moves ?
Post #163 Posted: Sun Oct 25, 2020 7:40 am 
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Bill Spight wrote:
If you don't want to be vague, let t = 0. But anyway, heuristics can help guide reading.

If you want 2 ≤ t < 3, then replace {8|5} with 6½ and play G1. If you are not more specific, what can be said about best play?
Gérard TAILLE wrote:
Bill Spight wrote:
Gérard TAILLE wrote:

OK Bill let's simplify the problem and formulate it in a different way.

Let's take G1 = {8|5||0} and G2 = {6|0}
and let's assume G3,G4,.. only of the forms {x|y} (I mean only simple gote) with g1 ≥ g2 ≥ g3 ≥ g4 ...
In addition let's suppose the game black to play G1 + G2 + G3 + G4 + ... is quite close and let's suppose that the god play allows black to win the game.
Question : what is the best way to proceed in order to be sure at 100% to choose as a first black move a winning move?


OK. Without loss of generality, let Gi = {gi|-gi}. Let gi' = gi - g(i+1) + g(i+2) - ....

I. g3 ≤ 1½
A. Black to play
1. G1 first: Score1 = 6½ - 0 + 1½ - g3' = 8 - g3'
2. G2 first: Score2 = 6 - 0 + g3' = 6 + g3'
Diff = Score1 - Score2 = 2 - 2g3'
-1 ≤ Diff ≤ 2
B. White to play
1. G1 first: Score2 = 0 + 6 - g3' = 6 - g3'
1. G2 first: Score1 = 0 + 6½ - 1½ + g3' = 5 + g3'
Diff = Score2 - Score1 = -1 + 2g3' ; Both players try to maximize Diff
-1 ≤ Diff ≤ 2
II. g4 ≤ 1½ ≤ g3 ≤ 3
A. Black to play
1. G1 first: Score1 = 6½ - 0 + g3 - 1½ + g4' = 5 + g3 + g4'
2. G2 first: Score2 = 6 - 0 + g3 - g4' = 6 + g3 + g4'
Diff = -1 + 2g4'
-1 ≤ Diff ≤ 2
B. White to play
1. G1 first: Score1 = 0 + 6 - g3 + g4' = 6 - g3 + g4'
2. G2 first: Score2 = 0 + 6½ - g3 + 1½ - g4' = 8 - g3 - g4'
Diff = 2 - 2g4'
-1 ≤ Diff ≤ 2
III. g5 ≤ 1½ ≤ g4 ≤ g3 ≤ 3
A. Black first
1. G1 first: Score1 = 6½ + g3 - g4 + 1½ - g5' = 8 + g3 - g4 - g5'
2. G2 first: Score2 = 6 + g3 - g4 + g5'
Diff = 2 - 2g5'
B. White first
1. G1 first: Score1 = 6 - g3 + g4 - g5'
2. G2 first: Score1 = 5 - g3 + g4 + g5'
Diff = -1 + 2g5'
IV. g6 ≤ 1½ ≤ g5 ≤ g4 ≤ g3 ≤ 3
A. Black first
Diff = -1 + 2g6'
B. White first
Diff = 2 - 2g6'

Interesting pattern. :)

When g(2n+1) ≤ 1½ ≤ g(2n), Black to play plays in G1 when 2 ≥ 2g(2n+1)', and White to play plays in G1 when 2g(2n+1)' ≥ 1.

When g(2n) ≤ 1½ ≤ g(2n-1), Black to play plays in G1 when 2g(2n)' ≥ 1, and White to play plays in G1 when 2 ≥ 2g(2n)'.

I expect that you already knew that. :) It is part of CGT, but not thermography.


Oops maybe you missed the point. The problem was not to find the best move but only to find a winning move!
My own result is far simplier:
if the score 6 + g3' allows me to win the game then my first play will be in G2 otherwise I will play in G1.
Could you verify if I am wrong?


How do you find g3'?

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At some point, doesn't thinking have to go on?
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Visualize whirled peas.

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 Post subject: Re: How evaluate double sente moves ?
Post #164 Posted: Sun Oct 25, 2020 8:03 am 
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Bill Spight wrote:
How do you find g3'?

OK Bill, let's proceed.
Assume black plays G2 and white answer G1 the score of the game will be
G2 - G1 + G3 - G4 ... = 6 - 0 + g3'

Now let's assume 6+g3' allows to win.
In this case black is sure to win by playing G2 because:
1)if white answers in G1 the score 6+g3' allows black to win
2)if white answers in G3 the exchange G2 - G3 is an ideal exchange for black, it couldn't be bad, and god told us black can win this game!

Now let's assume 6+g3' is a losing result for black.
Then black cannot play G2 because nothing can prevent white answering in G1
In this case it could not harm for black to try playing first in G1

Simple isn'it?

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 Post subject: Re: How evaluate double sente moves ?
Post #165 Posted: Sun Oct 25, 2020 8:20 am 
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Gérard TAILLE wrote:
Bill Spight wrote:
How do you find g3'?

OK Bill, let's proceed.
Assume black plays G2 and white answer G1 the score of the game will be
G2 - G1 + G3 - G4 ... = 6 - 0 + g3'

Now let's assume 6+g3' allows to win.
In this case black is sure to win by playing G2 because:
1)if white answers in G1 the score 6+g3' allows black to win
2)if white answers in G3 the exchange G2 - G3 is an ideal exchange for black, it couldn't be bad, and god told us black can win this game!

Now let's assume 6+g3' is a losing result for black.
Then black cannot play G2 because nothing can prevent white answering in G1
In this case it could not harm for black to try playing first in G1

Simple isn'it?


What is your aspiration?

Edit: And you did not answer how you find g3'.

Edit2: It's easier to find g16' than g3'. :)

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 Post subject: Re: How evaluate double sente moves ?
Post #166 Posted: Sun Oct 25, 2020 8:56 am 
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Bill Spight wrote:
Gérard TAILLE wrote:
Bill Spight wrote:
How do you find g3'?

OK Bill, let's proceed.
Assume black plays G2 and white answer G1 the score of the game will be
G2 - G1 + G3 - G4 ... = 6 - 0 + g3'

Now let's assume 6+g3' allows to win.
In this case black is sure to win by playing G2 because:
1)if white answers in G1 the score 6+g3' allows black to win
2)if white answers in G3 the exchange G2 - G3 is an ideal exchange for black, it couldn't be bad, and god told us black can win this game!

Now let's assume 6+g3' is a losing result for black.
Then black cannot play G2 because nothing can prevent white answering in G1
In this case it could not harm for black to try playing first in G1

Simple isn'it?


What is your aspiration?


The idea is only to offer the most efficient tool to win the game.

Let's take a middle game is which you hesitate between reducing peacefully the opponent territory and making a deep invasion.
My teacher explained me : try to count the game. If you are behing invade. What about if I am ahead? The answer is : invasion may be the best move from a theoritical point of view (god move) to reach the better score but, if your goal is only to win the reducing move will give you better chances.
It was a revelation for me. The goal of the go player is not to get the best score but to win. The ideal result is a sure win by 0.5 point isn't it?

For the endgame it is the same.
Imagine a very difficult area looking like
G1 = { ...?...|||||x|y} "|||||" being the root
and a very simple environment G2, G3, ... each Gi = {gi|-gi}

you do not need to analyse in details the area G1 in order to find the best move for the game.
Calculate only the score of the game by beginning by G2 - G1.
If the score is a winning one for black then play your first move in G2, otherwise play your first move in G1.
You see here the basic principle : if you see a simple way to win do not look for the best move. Otherwise choose the most difficult move.

The point here is to use the theory in a different way; to help go players to win the game rather than to play the best move.

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 Post subject: Re: How evaluate double sente moves ?
Post #167 Posted: Sun Oct 25, 2020 9:43 am 
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Bill Spight wrote:
How do you find g3'?

Gérard TAILLE wrote:
Bill Spight wrote:
Gérard TAILLE wrote:
OK Bill, let's proceed.
Assume black plays G2 and white answer G1 the score of the game will be
G2 - G1 + G3 - G4 ... = 6 - 0 + g3'

Now let's assume 6+g3' allows to win.
In this case black is sure to win by playing G2 because:
1)if white answers in G1 the score 6+g3' allows black to win
2)if white answers in G3 the exchange G2 - G3 is an ideal exchange for black, it couldn't be bad, and god told us black can win this game!

Now let's assume 6+g3' is a losing result for black.
Then black cannot play G2 because nothing can prevent white answering in G1
In this case it could not harm for black to try playing first in G1

Simple isn'it?


What is your aspiration?


The idea is only to offer the most efficient tool to win the game.

Let's take a middle game is which you hesitate between reducing peacefully the opponent territory and making a deep invasion.
My teacher explained me : try to count the game. If you are behing invade. What about if I am ahead? The answer is : invasion may be the best move from a theoritical point of view (god move) to reach the better score but, if your goal is only to win the reducing move will give you better chances.
It was a revelation for me. The goal of the go player is not to get the best score but to win. The ideal result is a sure win by 0.5 point isn't it?

For the endgame it is the same.
Imagine a very difficult area looking like
G1 = { ...?...|||||x|y} "|||||" being the root
and a very simple environment G2, G3, ... each Gi = {gi|-gi}

you do not need to analyse in details the area G1 in order to find the best move for the game.
Calculate only the score of the game by beginning by G2 - G1.
If the score is a winning one for black then play your first move in G2, otherwise play your first move in G1.
You see here the basic principle : if you see a simple way to win do not look for the best move. Otherwise choose the most difficult move.

The point here is to use the theory in a different way; to help go players to win the game rather than to play the best move.


There is a lot to untangle here.

First, in the context of this discussion, playing safe when you are ahead is hardly news. We both knew that beforehand, and I doubt if our readers were ignorant of that, either.

Second, brute force reading is not an application of theory.

Third, when the theory can tell you best play with less calculation, why not use it?

Fourth, when the theory tells you potential shortcuts that apply more generally than any one game, why not learn them?

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 Post subject: Re: How evaluate double sente moves ?
Post #168 Posted: Sun Oct 25, 2020 10:06 am 
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Bill Spight wrote:
First, in the context of this discussion, playing safe when you are ahead is hardly news. We both knew that beforehand, and I doubt if our readers were ignorant of that, either.

Sure

Bill Spight wrote:
Second, brute force reading is not an application of theory.

Yes

Bill Spight wrote:
Third, when the theory can tell you best play with less calculation, why not use it?

That is the point, the theory give us heuristically the best play but when a choice is difficult and the game very close you cannot avoid reading.

Bill Spight wrote:
Fourth, when the theory tells you potential shortcuts that apply more generally than any one game, why not learn them?


OC, I agree 100%.

The theory is perfect for finding the very very probable best move without brute reading. However it may be completed when a difficult choice arise in very close game. In that case (and only in this case) you cannot avoid reading and an help to show how to proceed may be interesting.

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 Post subject: Re: How evaluate double sente moves ?
Post #169 Posted: Sun Oct 25, 2020 1:58 pm 
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Click Here To Show Diagram Code
[go]$$W White to play
$$ --------------------
$$ | . . O . . . . . .|
$$ | . . X O . . . . .|
$$ | . . X O X . . O O|
$$ | X X X . X . . O .|
$$ | . X O O O O O O O|
$$ | X X . . . . . O .|
$$ | . X . . . . . O O|
$$ | X X . . . . . . .|
$$ | . . . . . . . . .|
$$ --------------------[/go]

G1 = { ...?...|||||x|y} "|||||" being the root
and a very simple environment G2, G3, ... each Gi = {gi|-gi}

Assume black to play and you hesitate between a play in G1 or G2.

With only the CGT theory you have to work hard in order to evaluate G1 but it may be too difficult (see Robert's post https://lifein19x19.com/viewtopic.php?p=260798#p260798) with various possiblities for the follow-up and with the risk of missing a tesuji). Maybe you will conclude the miai value is around ten points but it could be very easily be 9 or 11 if not more.
Certainly you hesitate between G1 and G2 because G2 is near from 10 it may even happen you have for G2 a very accurate value like 9⅞. How will you choose between a play in G1 and G2?

Why not trying my method (only in such situation of course)
You calculate the score of the sequence G2 - G1 + G3 - G4 ...
Here you have two solutions
1) you consider the environment G2 + G3 + G4 + ... as ideal and you take as a score the value:
G2 - G1 + G3 - G4 ... ≃ (x+y)2 + g2 + g3' ≃ (x+y)/2 + 3g1/2
if this score looks like a winning score you play in G2 but when the temperature drops and (x+y)/2 + 3gi/2 becomes a losing score then you play in G1
2) you do not trust that environment G2 + G3 + G4 + ... is ideal => you read the game for the real score:
g2 - g3 + ... + (x or y) + ... and if this score is a winning one then you play in G2 and you are sure to win, otherwise you play in G1 and your are sure it cannot be a bad move.

You have to understand my method is not aiming at replacing the CGT one. It would be really completly stupid. The intention is only to add a tool allowing the player to find the best move in the case the CGT evaluation is too difficult to calculate for a given area while it is easy for all the other areas.

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Post #170 Posted: Mon Oct 26, 2020 7:58 am 
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Gérard TAILLE wrote:
Bill Spight wrote:
Third, when the theory can tell you best play with less calculation, why not use it?

That is the point, the theory give us heuristically the best play but when a choice is difficult and the game very close you cannot avoid reading.


Actually, the theory gives exact results that enable us to restrict reading. Not in all cases, OC, but very often. The problem you just proposed is a case in point. With one exception, you do not have to read the whole branch of the game tree to find the best play. But with your proposed solution, you do have to read the whole branch to find out if one option wins the game. OC, if it does, so will best play. And best play in your problem will take less reading to find, as a rule.

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Post #171 Posted: Mon Oct 26, 2020 9:28 am 
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Bill Spight wrote:
Gérard TAILLE wrote:
Bill Spight wrote:
Third, when the theory can tell you best play with less calculation, why not use it?

That is the point, the theory give us heuristically the best play but when a choice is difficult and the game very close you cannot avoid reading.


Actually, the theory gives exact results that enable us to restrict reading. Not in all cases, OC, but very often. The problem you just proposed is a case in point. With one exception, you do not have to read the whole branch of the game tree to find the best play. But with your proposed solution, you do have to read the whole branch to find out if one option wins the game. OC, if it does, so will best play. And best play in your problem will take less reading to find, as a rule.


If the estimations of the two best areas are not close OC theory will give the best move, otherwise the probability to find the best move is near from 50%.
Take again the two areas G1 = {8|5||0} and G2 = {6|0}. The miai values are 6¼ and 6. The problem is that ¼ is already too small.
Now add the quasi ideal environment {5½, 5, 4½, 4, 3½, 3, 2½, 2, 1½, 1, ½}
By choosing G1 your score will be 8, and by choosing G2 your score will be 9.
It is quite logical : the result of the game is a naturel number (no decimals). That means that a very small change in the environment will easily add or substract a full 1 point. As a consequence when the miai values of G1 and G2 are close you hardly can expect to make the best choice with more than 50%.
In that case reading is the only possiblity and my proposal is just to choose the easiest reading!

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Post #172 Posted: Mon Oct 26, 2020 12:11 pm 
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Gérard TAILLE wrote:
Bill Spight wrote:
Third, when the theory can tell you best play with less calculation, why not use it?

That is the point, the theory give us heuristically the best play but when a choice is difficult and the game very close you cannot avoid reading.

Gérard TAILLE wrote:
Bill Spight wrote:
Actually, the theory gives exact results that enable us to restrict reading. Not in all cases, OC, but very often. The problem you just proposed is a case in point. With one exception, you do not have to read the whole branch of the game tree to find the best play. But with your proposed solution, you do have to read the whole branch to find out if one option wins the game. OC, if it does, so will best play. And best play in your problem will take less reading to find, as a rule.


If the estimations of the two best areas are not close OC theory will give the best move, otherwise the probability to find the best move is near from 50%.
Take again the two areas G1 = {8|5||0} and G2 = {6|0}. The miai values are 6¼ and 6. The problem is that ¼ is already too small.
Now add the quasi ideal environment {5½, 5, 4½, 4, 3½, 3, 2½, 2, 1½, 1, ½}
By choosing G1 your score will be 8, and by choosing G2 your score will be 9.
It is quite logical : the result of the game is a naturel number (no decimals). That means that a very small change in the environment will easily add or substract a full 1 point. As a consequence when the miai values of G1 and G2 are close you hardly can expect to make the best choice with more than 50%.
In that case reading is the only possiblity and my proposal is just to choose the easiest reading!


What do you mean by the quasi ideal environment? Does 5½ = {5½|-5½} or {2¾|-2¾}? My guess is the latter, since otherwise the choice will be in {5½|-5½}, hands down. Also, who plays first? OK, I went back and found that you had originally stipulated Black to play. Fine.

If 5½ = {5½|-5½} then the result of playing in the environment alone is 5½ - 5 + 4½ - . . . + ½ = 3, and that fits with your claim that starting with G2 gives a result of 9. But in that case your play is in the environment, not G2. If 5½ = {2¾|-2¾} then the result of playing in the environment is 1½ and the result of starting with G2 gives a result of only 7½. For ease of calculation we can write that environment as {5½|0} + {0|-5} + . . . + {½|0}. Doing so adds 1½ to the mean value of the environment and makes the result of starting with G2 9 points for Black. We find that result by calculating

6 - 0 + 5½ - 5 + 4½ - 4 + 3½ - 3 + 2½ - 2 + 1½ - 1 + ½ = 9

If Black plays in G1 we get {8|5} - 0 + 5½ - .... Note that {8|5} + {0|-3} = 5, so we can write the result as

5 - 0 + 5½ - 5 + 4½ - 4 + 3½ - 0 + 0 - 0 + 0 - 0 = 9½

That does not fit your calculations, so I do not know what you meant.

Edited for correctness.

Edit2: How about the environment {2½|-2½} + {2|-2} + (1½|-1½} + {1|-1} + {½|-½}? I don't know the answer, but let's give it a try.

If Black plays is G2 she gets this result:

6 - 0 + 2½ - 2 + 1½ - 1 + ½ = 7½

{8|5} + (1½|-1½} = 6½ so if Black plays in G1 she gets this result:

6½ - 0 + 2½ - 2 + 1 - ½ = 7½

It does not matter, does it? :lol:

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Post #173 Posted: Mon Oct 26, 2020 2:26 pm 
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Bill Spight wrote:
What do you mean by the quasi ideal environment? Does 5½ = {5½|-5½} or {2¾|-2¾}? My guess is the latter, since otherwise the choice will be in {5½|-5½}, hands down. Also, who plays first? OK, I went back and found that you had originally stipulated Black to play. Fine.

If 5½ = {5½|-5½} then the result of playing in the environment alone is 5½ - 5 + 4½ - . . . + ½ = 3, and that fits with your claim that starting with G2 gives a result of 9. But in that case your play is in the environment, not G2. If 5½ = {2¾|-2¾} then the result of playing in the environment is 1½ and the result of starting with G2 gives a result of only 7½. For ease of calculation we can write that environment as {5½|0} + {0|-5} + . . . + {½|0}. Doing so adds 1½ to the mean value of the environment and makes the result of starting with G2 9 points for Black. We find that result by calculating

OK Bill surely my calculation was not correct.

Anyway it is not the point and it seems there is here a misunderstanding.
My proposal is not intented to be an alternative to the current CGT theory. It is only another tool you can use if you face some difficulties for applying the CGT theory. I see two cases:

1)We saw in various previous posts that the best estimation is not always the best move. It might happen in particular when the environment is not enough "ideal" (I mean more or less miai or more or less tedomari). In that case when you hesitate between two moves with close miai values you may want to read the game and a method to read the game may be useful

2)The second case is far more important than the first case. If G1 = { ...?...|||||x|y} has one branch (here the black one) too complex to be evaluated ( follow-up with various alternatives not easy to visualise) my method allows you to ignore this complex branch and find all the same the winning move if it exists.

Why are you so negative with my proposal? By allowing to manage finding the winning move in game where CGT theory is in difficulty my proposal erase some drawbacks of CGT theory and as a consequence this CGT theory becomes more and more interesting doesn'it?

When you answered
Bill Spight wrote:
First, in the context of this discussion, playing safe when you are ahead is hardly news. We both knew that beforehand, and I doubt if our readers were ignorant of that, either.

That was obviously the answer of the go player but not the answer of the CGT theorician. If I understand correctly, CGT theory concern is only to find the best move and not simply to find a winning move. BTW the only way to know if a game is a winning one is to compare the score of the game to the komi. CGT theory is built to evaluate scores but does the theory compare this score with the komi?
My proposal is only a way to use the komi in order to find a winning move in situation where CGT theory faces difficulty to identify the best move.

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Post #174 Posted: Mon Oct 26, 2020 3:08 pm 
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Gérard TAILLE wrote:
Bill Spight wrote:
What do you mean by the quasi ideal environment? Does 5½ = {5½|-5½} or {2¾|-2¾}? My guess is the latter, since otherwise the choice will be in {5½|-5½}, hands down. Also, who plays first? OK, I went back and found that you had originally stipulated Black to play. Fine.

If 5½ = {5½|-5½} then the result of playing in the environment alone is 5½ - 5 + 4½ - . . . + ½ = 3, and that fits with your claim that starting with G2 gives a result of 9. But in that case your play is in the environment, not G2. If 5½ = {2¾|-2¾} then the result of playing in the environment is 1½ and the result of starting with G2 gives a result of only 7½. For ease of calculation we can write that environment as {5½|0} + {0|-5} + . . . + {½|0}. Doing so adds 1½ to the mean value of the environment and makes the result of starting with G2 9 points for Black. We find that result by calculating

OK Bill surely my calculation was not correct.

Anyway it is not the point and it seems there is here a misunderstanding.
My proposal is not intented to be an alternative to the current CGT theory. It is only another tool you can use if you face some difficulties for applying the CGT theory.


But it is intended to be superior to CGT in the example you had in mind, {8|5||0} + {6|0} + {x1|y1} + {x2|y2} + . . . + {xn|yn} + C, where xi and yi are numbers such that xi > yi, and C is a numerical constant, such that the result of Black playing first in {6|0} and otherwise (Edit: both sides) playing optimally is greater than 0.

It seems like your first attempt did not get the environment right, but I am sure that you can come up with a good one. :)

Quote:
Why are you so negative with my proposal?


Come up with a good environment for your example, and we'll talk about it. :)

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Post #175 Posted: Tue Oct 27, 2020 4:04 am 
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Bill Spight wrote:
Gérard TAILLE wrote:
Bill Spight wrote:
What do you mean by the quasi ideal environment? Does 5½ = {5½|-5½} or {2¾|-2¾}? My guess is the latter, since otherwise the choice will be in {5½|-5½}, hands down. Also, who plays first? OK, I went back and found that you had originally stipulated Black to play. Fine.

If 5½ = {5½|-5½} then the result of playing in the environment alone is 5½ - 5 + 4½ - . . . + ½ = 3, and that fits with your claim that starting with G2 gives a result of 9. But in that case your play is in the environment, not G2. If 5½ = {2¾|-2¾} then the result of playing in the environment is 1½ and the result of starting with G2 gives a result of only 7½. For ease of calculation we can write that environment as {5½|0} + {0|-5} + . . . + {½|0}. Doing so adds 1½ to the mean value of the environment and makes the result of starting with G2 9 points for Black. We find that result by calculating

OK Bill surely my calculation was not correct.

Anyway it is not the point and it seems there is here a misunderstanding.
My proposal is not intented to be an alternative to the current CGT theory. It is only another tool you can use if you face some difficulties for applying the CGT theory.


But it is intended to be superior to CGT in the example you had in mind, {8|5||0} + {6|0} + {x1|y1} + {x2|y2} + . . . + {xn|yn} + C, where xi and yi are numbers such that xi > yi, and C is a numerical constant, such that the result of Black playing first in {6|0} and otherwise (Edit: both sides) playing optimally is greater than 0.

It seems like your first attempt did not get the environment right, but I am sure that you can come up with a good one. :)

Quote:
Why are you so negative with my proposal?


Come up with a good environment for your example, and we'll talk about it. :)


Here are two examples
{8|5||0} + {6|0} + {5|0} + {4|0} + {2½|0} tedomari situation under t= 3
{8|5||0} + {6|0} + {5|0} + {4|0} + {3|0} + {½|0} miai situation under t= 3

But you know perfectly that Bill, because the theory itself tells you.
When you compare G1 ={8|5||0} and G2 = {6|0} theory tells you that the miaiValue(G1) = 3¼ > miaiValue(G2) = 3 and tells you that G1 is the best move with a high probability.
But the same theory gives you also a warning: G1 and G2 are incomparable => depending on the environment G2 may be better than G1.
No doubt that the incertainty grows when miaiValue(G1) and miaiValue(G2) becomes closer and closer to reach a complete incertainty in case of equality.

BTW it is really a strength of the theory to be able to give such warning! If you are a quite strong player you may in this case try to complete your analysis by some reading but without a method it could quite difficult. That is the point. It is really simply irrelevant to compare the two methods. In my mind my method is intended to reinforce the theory; in no way my method is an alternative; I see it as a complementary tool.

Bill, be sure that if I would not have adhered at 100% to the theory I would have only showed the drawbacks without searching a solution to diminish these drawbacks and I would have stopped quickly studying it!

Do you need a practical exemple to illustrate my second case where G1 = { ...?...|||||x|y} has a black branch quite difficult to evaluate?


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Post #176 Posted: Tue Oct 27, 2020 4:59 am 
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Bill Spight wrote:
Quote:
Why are you so negative with my proposal?


Come up with a good environment for your example, and we'll talk about it. :)


Gérard TAILLE wrote:
Bill Spight wrote:
Gérard TAILLE wrote:
Here are two examples
{8|5||0} + {6|0} + {5|0} + {4|0} + {2½|0} tedomari situation under t= 3
{8|5||0} + {6|0} + {5|0} + {4|0} + {3|0} + {½|0} miai situation under t= 3


OK, example 1. The result when Black starts in {6|0} is this:

6 - 0 + 5 - 0 + 2½ = 13½

And the result when Black starts in {8|5||0} is this:

(8 - 0) - 0 + 5 - 0 = 13

So Black should start in {6|0}. :)

Example 2.

The result when Black starts in {6|0} is this:

6 - 0 + 5 - 0 + 3 - 0 = 14

And the result when Black starts in {8|5||0} is this:

(8 - 0) - 0 + 5 - 0 + ½ = 13½

So Black should start in {6|0}. :)

(Above edited for correctness and clarity.)

What I derived in my original reply to you about this question says that to decide whether Black should play in {6|0} or {8|5||0}, we may ignore the hottest 2n positions in the environment of simple gote before the temperature drops below the local temperature of {8|5}, 1½.

So instead of reading and calculating the result of {8|5||0} + {6|0} + {5|0} + {4|0} + {2½|0} we can ignore {5|0} and {4|0} and use {8|5||0} + {6|0} + {2½|0}.

Similarly instead of {8|5||0} + {6|0} + {5|0} + {4|0} + {3|0} + {½|0} we can use {8|5||0} + {6|0} + {3|0} + {½|0}.

As I said, for this comparison, CGT allows us to do less reading and less calculation. :)

Now, since we are ignoring only 2 positions, if all we are asking is if starting in {6|0} allows Black to win, and if not, let Black start in {8|5||0}, simply doing that will be easier in example 2, with only 5 additions and subtractions instead of 6. But, as I pointed out finding g3 - g4 + g5 - ... is easier than finding g16 - g17 + g18 - .... That was a point that you simply ignored. :(

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Post #177 Posted: Tue Oct 27, 2020 5:41 am 
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Bill Spight wrote:
Now, since we are ignoring only 2 positions, if all we are asking is if starting in {6|0} allows Black to win, and if not, let Black start in {8|5||0}, simply doing that will be easier. But, as I pointed out finding g3 - g4 + g5 - ... is easier than finding g16 - g17 + g18 - .... That was a point that you simply ignored. :(


Sure Bill you can. OC you can also improve my proposal!
I already saw clearly your point and you have here two choices:
1) you calculate g3 - g4 + g5 - ...
2) you avoid calculating g3 - g4 + g5 - ..., you look only at g3, g4, g5 ... to know if the number of gote is odd or even. Then you calculate g16 - g17 + g10 ... and you use the good formula to conclude.

Yes Bill I agree you can choose 2).
I far as I am concerned, as a go player, I prefer to calculate g3 - g4 + g5 - ... allowing me to know who wins the game! It is simplier to understand for a go player and the calculation can be often simplified by miai considerations.

OK Bill, no problem, if you prefer taking 2) option it is fine providing it doesn't matter for you to know if you will win the game.
If you are trained to read the yose OC you can choose 1). It is your choice. I only have my own preference.

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Post #178 Posted: Tue Oct 27, 2020 5:53 am 
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In CGT thermography yields heuristics, but CGT also finds exact results that simplify situations.

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 Post subject: Re: How evaluate double sente moves ?
Post #179 Posted: Tue Oct 27, 2020 8:30 am 
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Bill Spight wrote:
In CGT thermography yields heuristics, but CGT also finds exact results that simplify situations.


Now I do not see any more a misunderstanding on thermography heuristics.
In theory I suspect CGT finds exact results but the practice my be quite difficult

Remenber my post https://lifein19x19.com/viewtopic.php?p=260962#p260962

Assume G1 = { ...?...|||||x|y} "|||||" being the root, and a very simple environment G2, G3, ... each Gi = {gi|-gi}

Click Here To Show Diagram Code
[go]$$B
$$ --------------------
$$ | . . O . . . . . .|
$$ | . . X O . . . . .|
$$ | . . X O X . . O O|
$$ | X X X a X . . O .|
$$ | . X O O O O O O O|
$$ | X X . . . . . O .|
$$ | . X . . . . . O O|
$$ | X X . . . . . . .|
$$ | . . . . . . . . .|
$$ --------------------[/go]


Assume black to play and you hesitate between a move in G1 or G2. How do you proceed to choose?

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Post #180 Posted: Tue Oct 27, 2020 9:24 am 
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Read variations as far as necessary, then for each major branch possibly apply theory.

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