What is the value of a move in a yose ko?

[kvasir]

Taking into account a ko threat as a game might be satisfactory but in practice I am pretty sure it would be far too complex.

We agree on this. I just think some questions may not have real answers that are also simple.

How do you formalize the corresponding tree?

This depends on what question you want to answer.

One could try the following but it likely fails in some cases.

A = { 1, A │ 2, B }
B = { 2 │ -A - 2}
-A = { -2, -B │ -A, -1 }
-B = { 2 + A │ -2 }

Note that
A_0 = 2
since
B^L ≥ A
and
A_3 = { 1 - t │ t - 1 }

I'm not entirely sure what to say about this game. I'd tell you what the left and right walls of the thermography are and go above the freezing point (since that is where the action you are talking about is) but I only have vague idea how it should look.

I think actually writing out the moves made until local play stops (i.e. you have a number or a cold game), like I did in a previous post, is what is most useful when you don't know exactly what to expect. This makes many things explicit that would only be implicit and therefore helps with avoiding errors, it could also help with giving insights.

If you read the moves out from a graph or a tree representation isn't very important, I just think the tree notation is useful when it is concise enough. It is a taste.

What is my conclusion up to now?
In practice, providing I am not able to know which side will surely win the ko (or kos), I consider by default I will start a direct ko if t = (a-b)/3 and I will start a yose ko one move if t = (a-b)/4.
I know it is far from being ideal but it is simple. Do you think it is possible to propose a better advice in practice with the help of taxes?

I think t = (a-b) / 4 is right when the approaching player wants to force the capture. Why does he want to force the capture? Either he has advantage in ko threats or he is desperate

I'm less certain about the next one but I think t = (a-b) / 6 is reasonable but rather simplistic. The case that is being modeled is when the approaching player wants to delay the (almost) inevitable and get a trade. He wants this trade to happen in the direct ko. If he his successful in getting the approach move, then he may be able to claw back something in the hotter direct ko. As with all trades it is likely to pay off to have exact variations in mind but it can be too difficult.

I think I'd recommend in practice to start the ko earlier. The threat is to end up having played as-if t = (a-b) / 4. The other player is the one that has to defend against that.

When should the player that doesn't need an approach ko prevent the ko? I think before the other player should try to do so as late as possible. That means just before the other player would start it.

The range (a - b) / 4 >= t >= (a - b) / 6 should be interesting.

Anyway that is what I think and I haven't reviewed the previous posts yet

[Gérard TAILLE]

I tried to use your tax approach to analyse the approach ko situation. It is quite interesting indeed.
Assuming d_1 = d_2 (the tax for black or white are equal) then my result is that you have to play in the approach ko as soon as t = (a-b)/5.
Under which assumption did you reach the value t = (a-b)/6 ?

[kvasir]

I tried to use your tax approach to analyse the approach ko situation. It is quite interesting indeed.
Assuming d_1 = d_2 (the tax for black or white are equal) then my result is that you have to play in the approach ko as soon as t = (a-b)/5.
Under which assumption did you reach the value t = (a-b)/6 ?

I noticed the following error before and wanted to double check it.

Which we can rewrite as expressions for the freezing point t
t ≥ d_1
t ≥ d_3-d_2
t ≥ (a - b) / 2 - d_2 / 2 - d_4 / 2

If we assume that t=d_1=d_2, then the freezing point is given by the equation
t = (a - b) / 3 - d_4 / 2
but what about d_3 and d_4?

Now I think this equation
t = (a - b) / 3 - d_4 / 2
should have been
t = (a - b) / 3 - d_4 / 3

That is -d_4 / 3 instead of -d_4 / 2.

Realizing that the freezing point of G is in fact lower than that of H we instead assume
t(G)=d_1=d_2
t(H)=d_3=d_4

Now we can find a relationship between t(G) and t(H) using
t = d_3-d_2
so
t(G) = 1/2 t(H)
and
t(G) = ( a - b ) / 6
t(H) = ( a - b ) / 3

The conclusions, using the corrected equation above, should instead be
t(G) = ( a - b ) / 5
t(H) = ( 2 / 5 ) ( a - b )

I was confusing temperatures when I wrote the assumption
t(H)=d_3=d_4
it should have been be
d_3 = d_4
and this is not same as the freezing point t(H).

I tried to check this and it seemed to be correct for the freezing point (and the assumptions of the model and that d_2=d_3 and d_4=d_5). That is, the choice between ignoring the second ko threat or allowing right to capture is equally good, and starting the approach ko and not starting it are also equally good. That is what freezing point means.

Did this answer your question?

[Gérard TAILLE]

I tried to use your tax approach to analyse the approach ko situation. It is quite interesting indeed.
Assuming d_1 = d_2 (the tax for black or white are equal) then my result is that you have to play in the approach ko as soon as t = (a-b)/5.
Under which assumption did you reach the value t = (a-b)/6 ?

I noticed the following error before and wanted to double check it.

Which we can rewrite as expressions for the freezing point t
t ≥ d_1
t ≥ d_3-d_2
t ≥ (a - b) / 2 - d_2 / 2 - d_4 / 2

If we assume that t=d_1=d_2, then the freezing point is given by the equation
t = (a - b) / 3 - d_4 / 2
but what about d_3 and d_4?

Now I think this equation
t = (a - b) / 3 - d_4 / 2
should have been
t = (a - b) / 3 - d_4 / 3

That is -d_4 / 3 instead of -d_4 / 2.

Realizing that the freezing point of G is in fact lower than that of H we instead assume
t(G)=d_1=d_2
t(H)=d_3=d_4

Now we can find a relationship between t(G) and t(H) using
t = d_3-d_2
so
t(G) = 1/2 t(H)
and
t(G) = ( a - b ) / 6
t(H) = ( a - b ) / 3

The conclusions, using the corrected equation above, should instead be
t(G) = ( a - b ) / 5
t(H) = ( 2 / 5 ) ( a - b )

I was confusing temperatures when I wrote the assumption
t(H)=d_3=d_4
it should have been be
d_3 = d_4
and this is not same as the freezing point t(H).

I tried to check this and it seemed to be correct for the freezing point (and the assumptions of the model and that d_2=d_3 and d_4=d_5). That is, the choice between ignoring the second ko threat or allowing right to capture is equally good, and starting the approach ko and not starting it are also equally good. That is what freezing point means.

Did this answer your question?

Is I said in a previous post (https://lifein19x19.com/viewtopic.php?p=280303#p280303) I do not see how white can ignore two times a black ko threat. As a consequence I do not see how d_4 can exist. For the same reason I do not see how d_3 can also appear.

Curiously however I have the same result for t(G) = (a-b)/5
Let me explain my own calculation (I assume the execution of a ko threat allow to gain 2d)
Left
G => a - t
Right : white takes the ko to continue by a approach move
white wins the ko G => b + t + 2d_2
black wins the ko G => a + 2t - 2d_1
Assuming d_1=d_2=d The tree counts above are
C1 = a-t
C2 = b+t+2d
C3 = a+2t-2d
The freezing point correspond to C1=C2=C3
C1=C3 => a-t = a+2t-2d => 3t=2d
C2=C3 => b+t+2d = a+2t-2d => b+t+3t=a+2t-3t => t = (a-b)/5

Concerning position H you say t(H) = ( 2 / 5 ) ( a - b )
If I understand correctly H is a direct ko. In that case why don't we have t(H) = ( a - b )/3 ?

[kvasir]

Is I said in a previous post (https://lifein19x19.com/viewtopic.php?p=280303#p280303) I do not see how white can ignore two times a black ko threat. As a consequence I do not see how d_4 can exist. For the same reason I do not see how d_3 can also appear.

Curiously however I have the same result for t(G) = (a-b)/5
[...]

With my assumption if we have t = d_1 = d_2 and d_3 = d_4 then we might want to simplify the game form.

Maybe it helps if I do just that and replace d_1 and d_2 with t and d_3 and d_4 with a new variable d.

Left:
G => a - t
Right:
G => … => a - d_1 = a - t
G => … => a - d_3 + d_2 = 2t + a - d
G => … => t + b + d_4 + d_2 = 2t + b + d

We get these constraints that represent different outcomes, which I shall try to explain briefly.

Right can't make the approach move (outcome A):
a - t >= a - t

Right can make the approach move but left ignores a ko treat (outcome B):
a - t >= t + a - d

Right can make the approach move and ignores a ko threat (outcome C):
a - t >= 2t + b + d

We can simplify and define a decision surface for B-C.
If the following conditions are true then left will choose to ignore a ko threat
d >= 2t
a - b >= 3t + d
a - b >= t + 2d

The freezing point of G is when outcome B and outcome C are equal. That is how I deducted
t = 1 / 5 (a - b)
d = 2 / 5 (a - b)


What is this new variable d?

It is like t but also something very different. t is a tax for playing a move in the game. d on the other hand appears to be a liability, something that we believe we can get if our ko threat is ignored.

I just put in some d variables but you have raised an interesting question.

How do we expect ko threats to add up when they are ignored?
I think we need an answer to this question.

[kvasir]

Concerning position H you say t(H) = ( 2 / 5 ) ( a - b )
If I understand correctly H is a direct ko. In that case why don't we have t(H) = ( a - b )/3 ?

Sorry, I keep writing t(H) when it isn't the freezing point of H. What I meant was a critical point of G, the one when the play in H changes.

I should have written
d_3 = d_4 = ( 2 / 5 ) ( a - b )

Which is a critical point of G which I conflated with the freezing point of H. The critical point occurs when playing H from G, which is the source of the confusion.

If the approaching player wishes to gain from making the approach move it appears they have to play H hotter than the freezing point. I think that is correct, they ignore a ko threat after all to get there. Since this doesn't apply to the defending player, the assumption I made that d_3 = d_4 is probably too simplistic.

[Gérard TAILLE]

Concerning position H you say t(H) = ( 2 / 5 ) ( a - b )
If I understand correctly H is a direct ko. In that case why don't we have t(H) = ( a - b )/3 ?

Sorry, I keep writing t(H) when it isn't the freezing point of H. What I meant was a critical point of G, the one when the play in H changes.

I should have written
d_3 = d_4 = ( 2 / 5 ) ( a - b )

Which is a critical point of G which I conflated with the freezing point of H. The critical point occurs when playing H from G, which is the source of the confusion.

If the approaching player wishes to gain from making the approach move it appears they have to play H hotter than the freezing point. I think that is correct, they ignore a ko threat after all to get there. Since this doesn't apply to the defending player, the assumption I made that d_3 = d_4 is probably too simplistic.

I do not understand why the approach player must ignore a ko threat in order to reach position H.

Assuming t=t(G) (=> t = (a-b)/5)

$$B :b5: tenuki :w6: tenuki $$ ------------------+ $$ . . . . . . O X .| $$ . . . . . . O X X| $$ . . . . . . O X 4| $$ . . . . . , X O 1| $$ . . . . . . X O O| $$ . . . . . . X . .| $$ . . . . . . X X X| $$ . . . . . . . . .| $$ . . . . . . . . .| $$ . . . . . , . . .| $$ . . . . . . . 3 2| $$ . . . . . . . . .| $$ . . . . . . . . .|

Click Here To Show Diagram Code

[go]$$B :b5: tenuki :w6: tenuki $$ ------------------+ $$ . . . . . . O X .| $$ . . . . . . O X X| $$ . . . . . . O X 4| $$ . . . . . , X O 1| $$ . . . . . . X O O| $$ . . . . . . X . .| $$ . . . . . . X X X| $$ . . . . . . . . .| $$ . . . . . . . . .| $$ . . . . . , . . .| $$ . . . . . . . 3 2| $$ . . . . . . . . .| $$ . . . . . . . . .|[/go]

is a ko threat
is the answer to the ko threat

Because t=t(G) the two tenuki moves and are correct aren't they?
After this sequence to we reach again the initial position G. What is the result? The result is that white has lost one ko threat for nothing.
For that reason I think playing a ko threat with is a mistake. I think white must simply play in the environment to gain t points.

[kvasir]

I do not understand why the approach player must ignore a ko threat in order to reach position H.

It is usual to talk about ko threats when you play ko.

Because t=t(G) the two tenuki moves and are correct aren't they?
After this sequence to we reach again the initial position G. What is the result? The result is that white has lost one ko threat for nothing.
For that reason I think playing a ko threat with is a mistake. I think white must simply play in the environment to gain t points.

This argument would apply to most games and it isn't valid.

For one thing it isn't clear why losing a ko threat matters. It is not like there is any other indirect ko at this temperature.

Another thing is that you haven't specified what this temperature t is.

Consider if there is one such move at temperature t in another part of the Go board and it is gote. If the approaching / attacking player plays in the approach ko and the defending player plays in the other component, then isn't the attacker about to play another move at this temperature in the approach ko? If the defending player successfully fights the ko then they will get to play the last move in the approach ko. In the first case there is an even number of moves left on this temperature and in the second there is an odd number, the second case is better for the defending player and it is not clear that it is a mistake to prevent the approach move.

Both players have some strategic choices to make when playing an approach ko. I specified which choice the attacking player was making (I wrote "If the approaching player wishes to gain from making the approach move..."), your example is one case when the attacking player doesn't succeed in making the approach move. It is nevertheless always the other player's strategic choice if and how to defend.

[Gérard TAILLE]

I do not understand why the approach player must ignore a ko threat in order to reach position H.

It is usual to talk about ko threats when you play ko.

Because t=t(G) the two tenuki moves and are correct aren't they?
After this sequence to we reach again the initial position G. What is the result? The result is that white has lost one ko threat for nothing.
For that reason I think playing a ko threat with is a mistake. I think white must simply play in the environment to gain t points.

This argument would apply to most games and it isn't valid.

For one thing it isn't clear why losing a ko threat matters. It is not like there is any other indirect ko at this temperature.

For sure ko handling is a difficult issue.
Let's take a direct ko to try and simplify the reasonning.

$$B $$ ------------------+ $$ . . . . . . O X .| $$ . . . . . . O X X| $$ . . . . . . O X O| $$ . . . . . , X O .| $$ . . . . . . X O O| $$ . . . . . . X . X| $$ . . . . . . X X X| $$ . . . . . . . . .| $$ . . . . . . . . .| $$ . . . . . . . . .|

Click Here To Show Diagram Code

[go]$$B $$ ------------------+ $$ . . . . . . O X .| $$ . . . . . . O X X| $$ . . . . . . O X O| $$ . . . . . , X O .| $$ . . . . . . X O O| $$ . . . . . . X . X| $$ . . . . . . X X X| $$ . . . . . . . . .| $$ . . . . . . . . .| $$ . . . . . . . . .|[/go]

Here you can see (a-b)=18.
When can we decide that a player will choose to play in that corner?
OC it depends on the exact configuration of the whole board including the existence of ko threats and the existence of other ko. In that sense no answer can be made to this question.
In practice, for an amateur, you can only give advice assuming an environment that is a kind of average of all possible environments. You will assume this environment has a temperature t, you will calculate a local temperature equal to 18/3 = 6, and you will formulate the following advice:
if t > 6 play tenuki (play in the environment)
if t < 6 play in the corner and fight the ko
if t = 6 it does not matter if you play in the corner or in the environment.

Now my question is the following. You are white. Assuming t = 6 and assuming black chooses to take the ko, will you use a ko threat to fight this ko or will you play simply in the environment to take t points without losing a ko threat for the future?

[kvasir]

Now my question is the following. You are white. Assuming t = 6 and assuming black chooses to take the ko, will you use a ko threat to fight this ko or will you play simply in the environment to take t points without losing a ko threat for the future?

This ko is a position with four states. Two of them have even number of moves remaining for each player and the other two have odd number of moves remaining. If the there are n other moves available then we are effectively playing the game *n or *(n+1) depending on if we are in an odd or even state of the ko. *n is the game of Nim and first player wins when n is odd and loses when n is even. When we make our move we should make sure to move to a game when n is even, so our opponent loses.

The right strategy isn't to play the highest temperature. Maybe that is where there is the most action, so to speak. You shouldn't just play where there is the most action without thinking if it improves your chances.

Preserving ko threats also isn't a good general advice. There are many such things in Go that can become all important in certain situations but really don't warrant much consideration otherwise. It is something you can think of when there is the possibility of a heavy ko. But don't handicap yourself by trying to have more ko threats left when the game is finished, you won't get to take them home

[kvasir]

Btw if n is even then our direct ko position is something akin to

K = { *n | { … | *n } }

In other words, the ko is well worth fighting over. We shouldn't waste our precious moves on the ko, we should only play ko threats!

At least that is how it appears to me.

[Gérard TAILLE]

Now my question is the following. You are white. Assuming t = 6 and assuming black chooses to take the ko, will you use a ko threat to fight this ko or will you play simply in the environment to take t points without losing a ko threat for the future?

This ko is a position with four states. Two of them have even number of moves remaining for each player and the other two have odd number of moves remaining. If the there are n other moves available then we are effectively playing the game *n or *(n+1) depending on if we are in an odd or even state of the ko. *n is the game of Nim and first player wins when n is odd and loses when n is even. When we make our move we should make sure to move to a game when n is even, so our opponent loses.

The right strategy isn't to play the highest temperature. Maybe that is where there is the most action, so to speak. You shouldn't just play where there is the most action without thinking if it improves your chances.

Preserving ko threats also isn't a good general advice. There are many such things in Go that can become all important in certain situations but really don't warrant much consideration otherwise. It is something you can think of when there is the possibility of a heavy ko. But don't handicap yourself by trying to have more ko threats left when the game is finished, you won't get to take them home

$$B $$ ------------------+ $$ . . . . . . O X .| $$ . . . . . . O X X| $$ . . . . . . O X O| $$ . . . . . , X O .| $$ . . . . . . X O O| $$ . . . . . . X . X| $$ . . . . . . X X X| $$ . . . . . . . . .| $$ . . . . . . . . .| $$ . . . . . . . . .|

Click Here To Show Diagram Code

[go]$$B $$ ------------------+ $$ . . . . . . O X .| $$ . . . . . . O X X| $$ . . . . . . O X O| $$ . . . . . , X O .| $$ . . . . . . X O O| $$ . . . . . . X . X| $$ . . . . . . X X X| $$ . . . . . . . . .| $$ . . . . . . . . .| $$ . . . . . . . . .|[/go]

I am a little lost kvasir.
When in the above position you try to find the best move (a local move in the corner or a tenuki in the environment) what is your assumption concerning the environment? An "ideal" environment? A "rich" environment? a Nim environment with n (even or odd) gote moves at temperature t?
With an "ideal" or "rich" environment we can evaluate a local temperature equal to (a-b)/3. With a Nim environment it seems quite different and I do see some cases where you must fight the ko even if t > (a-b)/3.
Do you mean that assuming a "ideal" or "rich" environment is not a good idea? For professionals I guess you are right but for amateurs isn't it the best approach to simplify the move value estimation?

[kvasir]

$$B $$ ------------------+ $$ . . . . . . O X .| $$ . . . . . . O X X| $$ . . . . . . O X O| $$ . . . . . , X O .| $$ . . . . . . X O O| $$ . . . . . . X . X| $$ . . . . . . X X X| $$ . . . . . . . . .| $$ . . . . . . . . .| $$ . . . . . . . . .|

Click Here To Show Diagram Code

[go]$$B $$ ------------------+ $$ . . . . . . O X .| $$ . . . . . . O X X| $$ . . . . . . O X O| $$ . . . . . , X O .| $$ . . . . . . X O O| $$ . . . . . . X . X| $$ . . . . . . X X X| $$ . . . . . . . . .| $$ . . . . . . . . .| $$ . . . . . . . . .|[/go]

I am a little lost kvasir.
When in the above position you try to find the best move (a local move in the corner or a tenuki in the environment) what is your assumption concerning the environment? An "ideal" environment? A "rich" environment? a Nim environment with n (even or odd) gote moves at temperature t?
With an "ideal" or "rich" environment we can evaluate a local temperature equal to (a-b)/3. With a Nim environment it seems quite different and I do see some cases where you must fight the ko even if t > (a-b)/3.

I was trying to answer your question. The answer was that it is not correct that ko threats shouldn't be used or even that the hottest move should be played. I provided a counter example to show that even if the hottest moves elsewhere are equally hot as in this direct ko, it is not the case that you should always avoid ko threats. The example shows that you might not want to play any normal moves.

In the above position I prefer to think of what are the intrinsic properties of this position. I don't want to assume much about ko threats. I think if you understand this ko very well then you will be better equipped to to recognize how to play it well in realistic game positions.

Do you mean that assuming a "ideal" or "rich" environment is not a good idea? For professionals I guess you are right but for amateurs isn't it the best approach to simplify the move value estimation?

I don't know much about the use of analyzing in ideal or rich environments. I recall that it has to do with how many moves are available and how they are distributed. That is if I recall correctly.

I'm increasingly of the opinion that knowing how to play a position, including the different ways to do so, is what is important, not move values. My experience as a Go player and my experience playing against and with KataGo increasingly strengthens my view.

When you know some ways to play a position then you can choose between the ways that you know. Direct comparison between handful of options is easier in practice than finding exact values and probably lot less error prone. It is not realistic to know everything about every position but you can work with what you know.

$$B $$ ------------------+ $$ . . . . . . O X .| $$ . . . . . . O X X| $$ . . . . . . O X O| $$ . . . . . , X O .| $$ . . . . . . X O O| $$ . . . . . . X . X| $$ . . . . . . X X X| $$ . . . . . . . . .| $$ . . . . . . . . .| $$ . . . . . . . . .|

Click Here To Show Diagram Code

[go]$$B $$ ------------------+ $$ . . . . . . O X .| $$ . . . . . . O X X| $$ . . . . . . O X O| $$ . . . . . , X O .| $$ . . . . . . X O O| $$ . . . . . . X . X| $$ . . . . . . X X X| $$ . . . . . . . . .| $$ . . . . . . . . .| $$ . . . . . . . . .|[/go]

I like to think about the ko position that you gave in terms of the cooled game. If I must write it down, then I much prefer writing the cooled game, possibly in a simplified form. For example

G_t = { v - t || t + { ...| t - v } }

This way makes it very clear that the freezing point is t = 2/3 v

When I try to write down the approach ko like this it turns out to be much trickier. Somehow it seems obvious to me that the freezing point should be half the ko threat in the direct ko. What was not obvious to me was that it wasn't just

t = 1/3 v and d = 2/3 v

and even though I did the derivation twice now I'm still not certain it is correct or useful to say that

t = 2/5 v and d = 4/5 v

or even what that exactly means

Of course if the ko is much smaller then the difference isn't worth considering. There will be other errors, including rounding errors, that will be larger if the ko is small enough.

[kvasir]

Let's look at this model for an approach ko with explicit ko threats again.

First I want to correct some errors I made.

The game form for the model with corrections marked in red ink:

G = { a | G^R }
G^L = { a │ { a - d_1, G │ … } }
G^R = { { … │ { G^L │ … }, H + d_2 } │ H}

H = { H^L | b }
H^L = { a │ { a - d_3, { … | H^R } │ … } }
H^R = { { … │ { H^L │ … }, b + d_4 } │ b }

Now we play this game to discover its structure:

Left:
G => a-t (1)

Right:
G => G^R+t => G^RL => {G^L-t│…}+t => G^L => G^LR-t => a-d_1 (2)
G => G^R+t => G^RL => {G^L-t│…}+t => G^L => G^LR-t => G …ko continues
G => G^R+t => G^RL => H+d_2+t => H^L+d_2 => t+H^LR+d_2 => a-d_3+d_2 (3)
G => G^R+t => G^RL => H+d_2+t => H^L+d_2 => t+H^LR+d_2 => {…│H^R }+d_2 => t+H^R+d_2 => H^RL+d_2 => t+b+d_4+d_2 (4)
G => G^R+t => G^RL => H+d_2+t => H^L+d_2 => t+H^LR+d_2 => {…│H^R }+d_2 => t+H^R+d_2=> {H^L│…}+d_2 => H^L+d_2 …ko continues
G => G^R+t => G^RL => H+d_2+t => H^L+d_2 => t+H^LR+d_2 => {…│H^R }+d_2 => t+H^R+d_2 => no move => 2t+d_2+b (5)

As before I want to analyze the approach ko with the simplest assumptions about which ko threats are made. These assumptions were that the fight to make the approach ko is played at the initial temperature and the direct ko can be at a different temperature, both ko fights are symmetric in the size of the threats used by each player.

To make this clearer I’ll use the following change in variables.

Let t = d_1 = d_2, d = d_3 = d_4 and v = a = -b.

I’ll write the cooled game directly in a simplified form (something close to a canonical form) based on the game as playout out above

G_t = { v - t │ t + { v - d, t + d - v │ … } }

The structure of this form is as shown with references to the variations above

G_t = { (1) │ t + { (3) - t, (4) - t │ … } }

I don’t have to include (2) since it is bypassed by right’s choice, and I don’t need to include (5) since it is bypassed by right’s choice since the game has become hotter. All the other stops are in G_t

Looking at this game it may become clear that the decision between (3) and (4) will be a critical point, this decision is based on if
v - d ≤ t + d - v
and that yields our first critical point
t_0 ≥ 2 v - 2 d
If v - d ≤ t + d - v, then left prefers to abandon the direct ko and this point to a critical point if (1) and (3) cool to a number. That would happens if
v - t ≤ 2 t + d - v
this yields the critical point
t_1 ≥ 2 / 3 v - 1 / 3 d

If v - d ≥ t + d - v, then left prefers to win the direct ko and this point to a critical if (1) and (4) cool to a number. That does happen when
v - t ≤ t + v - d
There is a critical point here (I'll will skip a number to make room for a missing critical point)
t_3 ≥ 1 / 2 d

The rest of the critical points will be discovered shortly when we look into the phase changes

For t_0
G_(2 v - 2 d) = { v - t │ t + { v - d, t + d - v │ … } } = { v - t │ 2 t + d - v } = 1 / 3 v + 1 / 3 d

And below t_0
G_(2 v - 2 d > t) = { v - t │ t + { v - d, t + d - v │ … } } = { v - t │ t + v - d }

If t ≥ 1 / 2 d
G_(2 v - 2 d > t) = v - 1 / 2 d
else
G_(2 v - 2 d > t) = { v - t │ t + v - d}

Therefore, a more insightful form of the cooled game is
If t ≥ 2 v - 2 d
G_t = 1 / 3 v + 1 / 3 d
if 2 v - 2 d > t ≥ v - 1 / 2 d
G_t = v - 1 / 2 d
else
G_t = { v - t │ t + v - d }

And the right subgame’s form is
if t ≥ 2 v - 2 d
G_t^R = t + d - v
else
d - v

Notice that this has little resemblance to a direct ko. It is in fact only the right side of the simplified form, and that may obscure something.

I’m not completely sure but I think the thermography of the position is (btw there is a picture below)

if t ≥ 2 v - 2 d
lw(G_t) = 1 / 3 v + 1 / 3 d
else
lw(G_t) = v -t

if t ≥ 2 / 3 v - 1 / 3 d
rw(G_t) = 1 / 3 v + 1 / 3 d
if 2 / 3 v - 1 / 3 d ≥ t ≥ v - 3 / 4 d
rw(G_t) = 2 t + d - v
if v - 3 / 4 d ≥ t ≥ 1 / 2 d
rw(G_t) = v - 1 / 2 d
else
rw(G_t) = t + v - d

The thermography appears to unveil a critical point that we didn’t know about
t_2 = 2 / 3 v - 1 / 3 d

That seems to be all for critical points.

approach ko.png (34.67 KiB) Viewed 13630 times

About the ko threat for the direct ko

Right can only gain by fighting to make the approach move and then winning the direct ko if
1 / 3 v + 1 / 3 d ≤ v - 1 / 2 d
or to put it differently
d = 4 / 5 v (6)
is a fair value for the threat in the direct ko.

We need to look outside of the simplified game for an interaction between d and t. From (2) and (3) we have that
v - t ≥ v - d + t

That means if (3) is chosen over (4) if
1/2 d ≥ t (7)
and if (2) is chosen over (4) if
v - t ≥ 2t - b + d
or to put it differently
2 / 3 v - 1 / 3 d ≥ t (8)
and using (6) we have
3/5 v ≥ t

Notice that (7) and (8) already appeared in the thermography. I’ll take the second derivation of (8) as some indication that the 2nd line of the right wall is correct.

What is happening?

Notice that the freezing point of this game appears to be t_2 = v - 3 / 4 d and the action that we see above that point is above freezing. Below freezing the action for the attacking player appears to start at t_3 = v - 1 / 2 d which is easy to confuse with the freezing point, but technically the position was active before.

The action below freezing is a fight to gain move elsewhere for the approaching / attacking player while the defending player must finish off the ko. The action above freezing is a fight to make the approach move and start the direct ko.

Before we got as some sort of a value for the ko threat in the direct ko
d = 4 / 5 v
as such it can only be relevant above freezing. If the attacking player wishes to gain from making the approach move, then he needs to keep this exchange rate in mind when judging the trade. However, this model obscures much of the details of the direct ko.

[Gérard TAILLE]

I was trying to answer your question. The answer was that it is not correct that ko threats shouldn't be used or even that the hottest move should be played. I provided a counter example to show that even if the hottest moves elsewhere are equally hot as in this direct ko, it is not the case that you should always avoid ko threats. The example shows that you might not want to play any normal moves.

In the above position I prefer to think of what are the intrinsic properties of this position. I don't want to assume much about ko threats. I think if you understand this ko very well then you will be better equipped to to recognize how to play it well in realistic game positions.

I don't know much about the use of analyzing in ideal or rich environments. I recall that it has to do with how many moves are available and how they are distributed. That is if I recall correctly.

I'm increasingly of the opinion that knowing how to play a position, including the different ways to do so, is what is important, not move values. My experience as a Go player and my experience playing against and with KataGo increasingly strengthens my view.

When you know some ways to play a position then you can choose between the ways that you know. Direct comparison between handful of options is easier in practice than finding exact values and probably lot less error prone. It is not realistic to know everything about every position but you can work with what you know.

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Click Here To Show Diagram Code

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I like to think about the ko position that you gave in terms of the cooled game. If I must write it down, then I much prefer writing the cooled game, possibly in a simplified form. For example

G_t = { v - t || t + { ...| t - v } }

This way makes it very clear that the freezing point is t = 2/3 v

When I try to write down the approach ko like this it turns out to be much trickier. Somehow it seems obvious to me that the freezing point should be half the ko threat in the direct ko. What was not obvious to me was that it wasn't just

t = 1/3 v and d = 2/3 v

and even though I did the derivation twice now I'm still not certain it is correct or useful to say that

t = 2/5 v and d = 4/5 v

or even what that exactly means

Of course if the ko is much smaller then the difference isn't worth considering. There will be other errors, including rounding errors, that will be larger if the ko is small enough.

I agree entirely with you Kvasir. The priority is to understand how to play a position. Here, with a first ko and then a direct ko it is difficult to have a clear understanding of the position. Playing a local move is very simple but choosing between a local move, a pass (in a cooled game) or a ko threat is not easy.