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)

Click Here To Show Diagram Code

[go]$$B tenuki tenuki
$$ ------------------+
$$ . . . . . . O X .|
$$ . . . . . . 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.

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



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.

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

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?

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

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.

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?

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.

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.

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.



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.

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.

Because my formula are a little different I think we have to clarify two points: firstly what is the value of a move in a ko threat area and secondly when is the game finished.

1)What is the value of a move in a ko threat area?
Let'me take again the following example of ko threat area:

Click Here To Show Diagram Code

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

After the white ko treat we reach the following position:

Click Here To Show Diagram Code

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

This position is very easy to evaluate. The tree is {2|-4} and the miai value is +3.
Taking your notation is it a good understanding to say here that d = +3 ?
IOW when white plays the ko threat white wins d points and if later white executes the ko threat then white will gain again d other points.
When playing a cooled game is it a good understanding that when white plays a ko threat white gains d-t points?

2)When is the game finished?
In the cooled game you consider, is it a good understanding to say that the game is finished when both players prefer to pass instead of playing locally or instead of playing in a ko threat area?

First of all you are very courageous to try and draw the thermograph because it is a big and boring job. Congratulation for that kvasir.

Now concerning the resulting thermograph something must be wrong:
taking the S axe I can see : v-d > 1/3v+1/3d => 4/3d < 2/3v => d < 1/2v
taking the t axis I can see : 2/3v-1/3d > v-3/4d => 5/12t > 1/3v => d > 4/5v
=> the two axes are not consistant.

Not boring but very frustrating.



I see what you mean. Thank you for pointing this out.

At first I thought it was my choice to instead of a real projection (v, d, s) -> (t, s) to offer two anamorphosis. One of the top half of the thermography where right can win the ko if d is large enough and another of the bottom half where right can never win the ko anyway. This wasn't very well thought out, in effect I just thought I was done when the two halves looked right. Now I realize that I must have mixed up some of the constraints, not sure how exactly, it might be easier for me to start from scratch than it is to correct.

For now I'm going to think about the direct ko instead.

The idea behind assigning a number was to start with something simpler. At least derive what this number should be, before considering the various types of ko threats.

My hunch is that ±d (which is {d|-d}) isn't it. When either side plays in ±d they immediately gain. Adding this move would also increases the temperature of the game to d. This would also not allow for answering the ko threat; it isn't a threat at all.

Some of my ideas so far for -d are
+_2d = { 0 ││ {0 │ -2d} }
+_({-100 | 2d}) = {0 │││ {0 ││ {-2d│100} } }
{ 0 │ -a + ±b } if a + b = 2d
±(d) + ±(d) = { {2d│0} ││ {0│-2d} }
but I haven't really checked them.

Maybe +_2d is what is most straightforward. I have 2d since replacing a number with a game will otherwise create a tedomari when d >> t. At least that is my hunch.



Not exactly but maybe in practice. The players are indifferent to games when they aren't sufficiently cooled and play becomes completely forced when it is cooled down to t=0. Which I always think breaks the "temperature" analogy, shouldn't the game freeze over when it is cooled more, not less But this is how it works.

Basically, there can be a pass but that would mean we should replace that position with a number in the cooled game.

As for cycles there is handwaving. We sort of know what we want, which is to compare play that ends in positions that we can give a value to. We can eliminate cycles by ignoring them but also need to be careful not to try to analyze positions where we would in fact get stuck in a cycle.

About ko threats.

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

This is basically the cooled direct ko with some d ko threats, except that I skip a cooling by t when d appears. You could say I use d - t rather than d. That is in order to get the value for d for G not some value that needs to have t added back.

Using d instead 2d appears closer to how I'd think about it. Making the ko threat often doesn't have a value far from zero and then the follow up has to make up everything, it is convenient to just get that value instead of half that value.

There are some cases that confuse this picture but I think my form is effectively that of both players having some +_d and -_d threats available.

The form is a simplification of the following overly complicated one (typos likely)
G_t = { v - t ││││││ t + { { … ││││ t + { { … │││ t + { v + { … │ -d } - t ││ … } } - t │││││ … }, t + { d │││││ … } - v } -t ││││││ … } }

Maybe one could write the direct ko in some way, say H and then add ko threats
G = H + n * +_(d) + m * -_(d)

One can play out this game without adding the components explicitly. I might do just that when if I feel energetic sometime but there is lot to be said about using the simplest forms one can find. For one thing they can match how one thinks of these positions when complicated trees, in contrast, are far from the human experience of playing Go. I think I'd much prefer writing down a spanning tree for the actual Go game and identify how to write a simple form based on that.

Yes kvasir it is a good idea to analyse first a "simple" (???) direct ko.

In order to show you my own analyse I will change some notation.

First of all I consider the two following positions I call G and H:

Click Here To Show Diagram Code

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

Position G

Click Here To Show Diagram Code

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

Position H

In addition I define position G' identical to G but in G' black is not allowed to take the ko.
Similarly I define position H' identical to H but in H' white is not allowed to take the ko.

I call K the swing value of the ko : here K = 18
I call t the temperature of the environment or, if you prefer, the cooled value.

Now I assume the fight of the ko should take place. That means K > 3t.
Finally I assume each player have the same big set of ko threats : 2d1 > 2d > 2d3 ... > 0

At the beginning of the ko fight each player is able to use big ko threats and the opponent has te answer the ko threats. As soon as the value of ko threat become small enough then one player will ignore ko threat and the opponent will end the ko.
What is the value 2δ of this freezing ko threat?
Starting from G
1) if black wins the ko then the count is C1 = K - 2δ
2) if white wins the ko then the count is C2 = 2δ - t
The freezing ko threat appears when C1 = C2 => 4δ = K+t

Now let define d by δ = t + d
Then C1 = C2 => 4(t+d) = K+t => K = 3t + 4d

Simple algebra will now give you the value of each move assuming the freezing ko threat.

Considering the sequence:

Click Here To Show Diagram Code

[go]$$B ko threat, answer to ko threat, ko threat, ko threat execution
$$ -----------------+
$$ . . . . . . X O .|
$$ . . . . . . X O O|
$$ . . . . . . X O 1|
$$ . . . . . , O X 4|
$$ . . . . . . O X X|
$$ . . . . . . O 6 O|
$$ . . . . . . O O O|
$$ . . . . . . . . .|
$$ . . . . . . . . .|
$$ . . . . . . . . .|[/go]

The count of G,G',H,H' as well as the value of each move are given in the following figure:

Comment 1: you can see that the value of is t+3d while the value of is only t+d. Why? Beause gains OC the value of the ko threat (t+d) but gains also point in the local position because before the local position is H' while after the local position is H.
Comment 2: all move values are greater than t; that means that neither player can play tenuki without losing points. Fighting the ko is mandatory.
Comment3: it is interesting to note that the sequence to is gote for black and black gains only t points (and not K/3). The reason for that is that white gets compensation for losing the ko.

OK, you define every variable differently than I have

I think
C2 = 2δ - t
should be
C2 = 2δ - K/3
since you said K > 3t.

I take your t to be the temperature of the environment excluding the ko.

So freezing of the ko is when
2δ = 1/3 K

Your 2d and 2δ correspond to my d but otherwise I think the analysis is the same.

I have often confused different temperature, I can only recommend to use different symbols for every different temperature and avoiding to consider different temperatures at the time same time.

I have preference against saying the temperature of the environment is T since I don't know what that means and it would seem to me if there is a ko position with a critical temperature close to T that the temperature of the two combined could be something that is not obvious.

Btw I like your graphs!

Actually I have more a detailed analysis of the direct ko that I want to spend some time double checking.

C2 is the final count when white wins the ko. This situation corresponds exactly to the example sequence to I showed.
After the sequence the count in the local position is 0 because black has lost the ko, and the count in the ko area is 2δ (one white move is for the ko threat and a second move for white executing the ko threat). Because the sequence is gote for black the count for the cooled game is
C2 = 2δ - t.

I do not see how you can have C2 = 2δ - K/3. When I say K > 3t that means the ko is active but the cooled game remains with t not K/3.

Looking at it closer I'm not sure I follow.

We start in G where black can take the ko to move to H' and white can just end it by capturing everything to move to C2. White can play a ko threat in H' but black can ignore it to move to C1. There is one ko threat here, black played one more move than white to get to C1 and white played one more move to get to C2. Therefore
C1 - d - t = C2 + t
using t the freezing point and d for some ko threat.

You seem to literally say that
C1 = C2