We show under another topic that evaluating a yose ko is not that easy.
Here I try to analyse a very simple yose ko.

Click Here To Show Diagram Code

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

the tree for evaluating the position is the following:

Assume no ko threats available. It seems to me that neither player will play in the corner until dame phase. That means that the temperature of the corner is t = 0. In this dame phase I expect either white plays the branch AC or black plays the branch AB BD DF FG for a count -9 in the corner (white kills the four black stones).

Now assume black has one ko threat. That means that black can choose the branch AB BD DE while white can OC play AC. Because the tally is here equal to 4 that means that the temperature of the position in the corner is t = (9 + 9)/4 = 4.5

I am not sure my reasonning is correct. What is your view?

If we write this game in the following form, then we do reach the conclusion that t = 0.

A={B│-9}
B={D│{A│-9} }
D={9│{B│-9} }

Let me show why (though you had the same argument but I want to contrast it).

First if left/black starts the game progresses as follows
A => B => {A│-9} => A => -9
and when right/white starts
A => -9

But this is only the case if there are no useful moves available. In that way it doesn't depend on ko threats as much as it depends on there being no other moves at all.

If we instead write the game using d_1 and d_2 to mean a tax corresponding to ignored ko threats
A={B│-9}
B={D│{A, D - d_1│-9} }
D={9│{B, 9 - d_2│-9} }
then the progression of the game is different when black starts.
A => B => {A, D - d_1 │ -9} => D - d_1 => {B, 9 - d_2 │ -9} - d_1 => 9 - d_2 - d_1

That means the left stop is now
9 - d_2 - d_1
which will be worthwhile to pursue.

I think it is useful to think of this position in the following way. The position becomes active earlier than one might think but black will need to pay prize money to salvage the group. It is in a way like a wrecked ship, it has lot less value for its owner than before but it is something relative, but as time passes it is not unlikely that someone will show interest in salvaging the wreck for a prize. So black like the wreck owner isn't keen on lunching an expensive salvage effort but there could be a bargain when the game draws to a close. Since black has to spend a move and allow white two free moves it can be understood that the bargain price is rather lower than the temperature at which the position becomes active.

I'm not very confident in my ability to analyze such positions rigorously yet so I'll leave that for another time.

I do not understand the tree under B: B={D│{A│-9} }
for me this tree is simply B = {D|A}

The notion of tax corresponding to ignored ko threats is interesting. Does it already exist a theory based on such tax? As far as I am concerned I need some time to build some examples allowing to understand what it means exactly.
At first sight I assume this tax will also exist for the other side : assuming white will kill the black group then white might also pay a tax. Is it a good understanding?

Maybe I misunderstood the tree. Anyway, I think the tree I gave wasn't exactly what I wished. Let me give you a simpler case below.



Usually the tax and ko threats are treated as the same thing. I recall seeing threats treated separately. This was done mostly to demonstrate that most of the time it is the same thing. I can't recall where I saw it, is very long time ago.

If you only want to get a minimum value for ko threats in a ko then you can use the tax. With that value you can start to reason about the ko as a Go player. But you seem to want to ask questions like what if there are no ko threats or even no other moves at all.

Let's consider the game that follows, which I believe is a simple ko that is over a difference of a - b points.

Usually I'd write

H = { a ││ {… │ b } }
H_t = { a - t ││ t + { …│ t + b } }

and quickly get

t(H) = (a - b) / 3

but let's include the ko threats in the game and try to write a form for the game that has some explicit ko threats.

Here we go

G = { a │││ { { … ││ { G │ … } , b + d } │ b } }
G_t = { a - t │││ t + { { … ││ t + { G - t │ … }, t + b + d } - t │ t + b } }

My intention here is that when it is illegal to take in the ko, due to the normal rules of Go, that they can offer instead a choice. The choice that is on offer is for the second player to end the ko at the cost of d or to allow the first player to return to the original position, where it must be the second player's turn to make a move. This is the normal rules of Go and I have used d since ko threat in Japanese is "kodate" which has a "d".

I find it to be is a bit tricky to get the right structure but I hope I managed. I tried to write a form for the approach ko but I became too confused

Let's analyze this game.

Now, if left is to play the game progresses as follows

G => a - t

and when right starts

G => G^R + t => G^RL => { G │ … } + t => G Mistake! b + d + t >= G and right should choose differently.
G => G^R + t => G^RL => b + d + t

btw. It is mostly irrelevant if we use the cooled game or not, but the "+ t" and "- t" are useful to indicate who has just played.

We equations, from the game progression above, for the freezing point and the mast value / midpoint.

t(G) = (a - b) / 2 - d / 2
m(G) = (a + b) / 2 + d / 2

If we require that d = t(G), then this becomes

t(G) = (a - b) / 3
m(G) = (2a + b) / 3

and this is same as if we had used tax (and more straightforward form for the game).

This is why it is usually not necessary to treat ko threats differently than a tax. It does also complicated the analysis, as can be seen by the more complex form of the game and the more complex formulas that result from the analysis of this most simple ko.

Let me try to amend my analysis above and then try to answer your question about the approach ko.

Earlier I wrote

but this this is one-sided. Only the possibility that left ignores a ko threat is considered. Now I want to allow right to ignore a ko threat, this will be important for the approach ko.

Here we go
G = { a | G^R }
G^L = { a ││ { a - d_1, G │ … } }
G^R = { { … ││ { G^L │ … }, b + d_2 } │││ b }

G is the game when left/black is one move from ending the ko for a value of a and there is no move disallowed. G^L is the same position as in G but now right/white is disallowed from capturing in the ko. G^R is the position when right/white has captured in the ko and left/black is disallowed from capturing back.

d_1 and d_2 are the values of ko threats. I'm using the convention, in this post, that odd number threats are made by right/white and even number threats by left/black. This helps with keeping track of which is which.

The cooled form of these games (when cooling by t) are
G_t = { a - t │ t + G_t^R }
G_t^L = { a - t ││ t + { a - d_1 - t, G - t │ … } }
G_t^R = { { … ││ t + { G^L - t │ … }, t + b + d_2 } - t │││ t + b }

Now let's play the cooled form

Left
G => a - t

Right
G => t + G^R => G^RL => t + b + d_2
G => t + G^R => G^RL => t + { G^L - t │ … } => G^L => t + G^LR => a - d_1
G => t + G^R => G^RL => t + { G^L - t │ … } => G^L => t + G^LR => G => … ko continues

As in the earlier post we have some equations (for when the game is active, or not a number) that can help us glean some insights
a - t ≥ t + b + d_2
a - t ≥ a - d_1

We obtain expression for t with some algebra and
t ≤ (a - b) / 2 - d_2 / 2
t ≤ d_1

The temperature that is of interest is the freezing point which is the highest temperature which the game is active. Therefore we consider only the quality
t = (a - b) / 2 - d_2 / 2
t = d_1

Now if we assume that t = d_1 = d_2 then we have what we expected
t = (a - b) / 3

Just as before.

Now I'll try to write down a similar game, one that makes ko threats explicit, for the approach ko.

This is only slightly disying so here we go
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 │ … } }
H^R = { { … │ { H^L │ … }, b + d_4 } │ b }

Let's break this down.

G, G^L and G^R are mostly as before but now right/white isn't fighting to end the ko but to get to make an approach move. You can see my earlier post for not much less briefer description of G.

H is the game when right/white is one move away from ending the ko for a score of b and there is no disallowed move in the ko. H^L is when left/black is one move away from ending the ko for a score of a and it is disallowed by the ko rule for right/white to take in the ko. H^R is the same position as H but now it is disallowed by the ko rule for left/black to take back in the ko.

Let's play the cooled form of this game.

Left:
G => a - t

Right:
G => G^R + t => G^RL => { G^L - t │ … } + t => G^L => G^LR - t => a - d_1
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
G => G^R + t => G^RL => H + d_2 + t => H^L + d_2 => t + H^LR + d_2 => H + d_2 => t + H^R + d_2 => H^RL + d_2 => t + b + d_4 + d_2
G => G^R + t => G^RL => H + d_2 + t => H^L + d_2 => t + H^LR + d_2 => H + d_2 => t + H^R + d_2 => H^RL + d_2 => t + { H^L │ … } + d_2 => H^L + d_2 …ko continues

Now we have three conditions for the freezing point
a - t ≥ a - d_1
a - t ≥ a - d_3 + d_2
a - t ≥ t + b + d_4 + d_2

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?

If we were to assume that t=d_3=d_4, which is incorrect in this case, then the freezing point would be
t = 2/9 (a-b)
but this is incorrect.

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

Note that
t(G) = 2/9 (a-b)
would be correct if t(G) >= t(H) but it is not the case, it is t(G) < t(H).

I hope this helps. Let me know if you spot some errors.

Basically, if a = 9 and b = -9 then on he assumption that ko threats are same as the temperature
t(G) = 3
t(H) = 6

There are many things to consider, one could play with the values of d, but as seen above it could be easy to miss it when G or H freezes.

I think I have understood the idea.
Considering the details of your analyse I have difficulties with the sequence
Right
G => t + G^R => G^RL => t + b + d_2
I would have written G^RR instead of G^RL. Maybe I have not quite understood what G^RL means?

Considering the sequence:
G => G^R + t => G^RL => H + d_2 + t => H^L + d_2 => t + H^LR + d_2 => H + d_2 => t + H^R + d_2 => H^RL + d_2 => t + b + d_4 + d_2
I do not understand how black can gain the two taxes d_2 and d_4.
I understood that a tax was paid only when the opponent do not answer to a potential ko threat. But in the sequence above black can use only one ko threat because black plays tenuki only 3 times. In order to gain one tax we need 2 consecutive tenuki and in order to gain two taxes then we need 2 times 2 consecutive tenuki.
As a consequence I think the result of the sequence is
2t + b + d_2 instead of t + b + d_4 + d_2
but I am not sure at all

G^RL is meant to mean "the left side of the right side of G" or maybe more clearly G^RL = (G^R)^L.

G => t + G^R => G^RL => t + b + d_2

This sequence is alternative play in the game when right starts. I can make it more explicit by putting it on different lines with explanations.

In G right moved to t + G^R.
In t + G^R left moved to (G^R)^L.
In (G^R)^L right moved to t + b + d_2.
t + b + d_2 is a terminal position and the game ends.

I probably should write "In G_t right moved to t + (G^R)_t ..." but this is so cumbersome. Alternatively I could do away with the "+ t" and "- t" and just add them afterwards.

The d factors are for ignoring threats that are played somewhere else. They end up taking the place of the t tax when we compared to the game were the ko threats are not part of the game. This is because adding the ko threats to the game means we don't skip any moves, and there can therefore only be 1, 0 or -1 of t.

The only case we might skip a move is if we come to a position/game with a lower freezing temperature. In those cases we should replace the game with the midpoint (OK, maybe there can be some interesting cases when a move increases the temperature by a lot but this is not the case for simple positions).

I just checked that sequence and it appears correct. I can try to give a diagram.

Click Here To Show Diagram Code

[go]$$B B7 at B1
$$ ------------------+
$$ . 5 . . . . O X .|
$$ . 6 . . . . O X X|
$$ . . . . . . O X 4|
$$ . . . . . , X O 1|
$$ . . . . . . X O O|
$$ . . . . . . X 9 3|
$$ . . . . . . X X X|
$$ . . . . . . . . .|
$$ . . . . . . . . .|
$$ . . . . . , . . .|
$$ . . . . . . . . 2|
$$ . . . . . . . . 8|
$$ . . . . . . . . .|[/go]

Notice that we have left=white and right=black in this example but let's try to pair up the moves in the diagram and the game progression.

Black takes in the ko.
G => G^R + t
White makes ko threat.
G^R + t => G^RL
Black ignores a ko threat and makes the approach move to go to H.
G^RL => H + d_2 + t
White takes back in the ko.
H + d_2 + t => H^L + d_2
Black makes a ko threat.
H^L + d_2 => t + H^LR + d_2
White responds to the ko threat.
t + H^LR + d_2 => H + d_2
Black takes the ko.
H + d_2 => t + H^R + d_2
White makes a ko threat.
t + H^R + d_2 => H^RL + d_2
Black ignores a ko threat and ends the ko.
H^RL + d_2 => t + b + d_4 + d_2

Black has played one move move in the ko which is +t and ignored two ko threats. Ignoring costs black d_4 and ignoring costs black d_2.

I have realized not that the t changes. At the root of the game it is t(G) but then when we move into H it is t(H). There need to be two variables, if we use only one variable then possibly this should be

2t + b + d_4 + d_2

since t(H) = 2 t(G).

I appear to have carried this error into the algebra. Which I should correct when I figure out what to do with the two t.

Another way to reason about this, and without introducing ko threats explicitly, is to notice that the approach move is sente and therefore the ko to get the approach move will behave like a gote move in G, then that the ko in H is a direct ko. That is we are fighting to have one out of three moves in the direct ko but this fight is a two move fight, therefore we divide the width by two times three which is six.

Oh, I see where is the issue.
I consider your sequence is not finished because at the end, the game is still at a high temperature due to the two ko threats used but not executed.
In order to return to the initial temperature t(G) we need to add two moves:

Click Here To Show Diagram Code

[go]$$B B7 at B1
$$ ------------------+
$$ . 5 . . . . O X .|
$$ . 6 . . . . O X X|
$$ . . . . . . O X 4|
$$ . . . . . , X O 1|
$$ . . . . . . X O O|
$$ . . . . . . X 9 3|
$$ . . . . . . X X X|
$$ . . . . . . . . .|
$$ . . . . . . . . .|
$$ . . . . . , . . .|
$$ . . . . . . . . 2|
$$ . . . . . . . . 8|
$$ . . . . . . . . .|[/go]

and then

Click Here To Show Diagram Code

[go]$$Bcm9
$$ ------------------+
$$ . O . . . . O X .|
$$ . X . . . . O X X|
$$ . . . . . . O X .|
$$ . . . . . , X . X|
$$ . . . . . . X . .|
$$ . . . . . . X X X|
$$ . . . . . . X X X|
$$ . . . . . . . . .|
$$ . . . . . . . . .|
$$ . . . . . , . . .|
$$ . . . . . . . 2 O|
$$ . . . . . . . 3 O|
$$ . . . . . . . . .|[/go]

execution of the white ko threat
answer to white ko threat

As you can see white has lost one ko threat.
A better sequence for white would be

Click Here To Show Diagram Code

[go]$$B B7 at B1
$$ ------------------+
$$ . 5 . . . . O X .|
$$ . 6 . . . . O X X|
$$ . . . . . . O X 4|
$$ . . . . . , X O 1|
$$ . . . . . . X O O|
$$ . . . . . . X 9 3|
$$ . . . . . . X X X|
$$ . . . . . . . . .|
$$ . . 2 . . . . . .|
$$ . . . . . , . . .|
$$ . . . . . . . 0 8|
$$ . . . . . . . . .|
$$ . . . . . . . . .|[/go]

Here is not a ko threat but just a move in the environment to gain t points.
is the execution of the ko threat

In any case the result I find is : 2t + b + d_2
The point is that white cannot execute two ko threats.

Assuming that d_2 is answered and only d_4 is a real ko threat has the problem that we are then playing a different game were d_2 and d_4 weren't numbers that we chose but hot games. You could analyze this by using something like D_i = { 2 d_i | 0 } instead of d_i. I think you will find that there will only be one t, two d_2 and no d_4.

In fact you could choose other forms for ko threats.

You may find that you are not only playing a more complex game but that you are playing multiple games at once. It needs to be clarified what each game means, it is to be expected that there are different conclusions when the games are different. Maybe there are two or three conclusion about how to play this approach ko that mirror something that is reasonably usual in actual play. I think the approach should be to clarify when different conclusions are useful for real Go games.

Let me offer a beginning of this analysis.

Only right can win the direct ko
If you wish to analyze the case when right (which is black in the diagram) could win the ko at will and left (which is white in the diagram) can't, then we could proceed by assuming that there is a fixed t from the environment and that t = d_1 = d_3 = d_2 = d_4. We do this to find what the temperature from the environment, excluding G, needs to be for right to reap the reward from being the only one able to fight the direct ko that results from winning the fight for the approach move.

In this case we should get t = (a - b) / 4, which I think we do if we use my game form. That conclusion of course hinges on left not having enough good threats for the direct ko, this can be is reasonable since we don't have left starting the ko. It is like a ko that waits until there aren't good any threats left and the question is when should left prevent a ko and when should right start it.

Only left can win the direct ko
If we assume the opposite, that only left can win the direct ko, then we must be careful about when right starts this ko. Left can end either ko were as right can only end the direct ko. If right he ignores a threat to get to the direct ko then this becomes a sunk cost. He'd end up playing H + d2, which is unfavorable unless -d_3 + d_2 >= d_1. But we must be careful here, our assumption seems too strong, it could imply d_1 < 0, that left will bribe right to play the direct ko.

We need to be careful when assuming who wins the first ko
It is actually all about the second ko. Only assuming something about this first ko seems incorrect, maybe left could always win both or right could always win both and that will influence the fight for the first ko but I think we should always start from assumptions about the direct ko.

Maybe you are right that is most useful to start and end in the same t from the environment. I need to think about. It is often not what happens in real Go games. Often the t from the environment appears to be greater after a ko fight, sometimes it is much lower. You also often fight small ko with normal moves that you'd wish to play anyway but large heavy ko with moves that you wouldn't play.

But maybe the t has to be assumed to be unchanged while play continues in the same game form. I'll try to review the above posts later.

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.
Let's take as an example a very common and simple (small) ko threat area:

Click Here To Show Diagram Code

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

Four continuations might happen
1) Neither players play in this area => at the end of the game the count is c = +2
2) White plays the ko threat and black answers => at the end of the game the count is c = +2
3) White plays the ko threat, black ignores and white take the two black stones => at the end of the game the count is c = -4
4) Black plays first in the area in order to suppress the white ko threat => at the end of the game the count is c = +1
How do you formalize the corresponding tree?
G = {+1||+2|-4} is almost satisfactory but the first case above does not appear here.
In addition you can also note that after continuation three, white has created a new ko threat for black.

Because continuation 4) looks uncommon I can show you the following example:

Click Here To Show Diagram Code

[go]$$W White to play
$$ | X X X X X X X O . |
$$ | . . X . O . X O O |
$$ | X X X X X O X O X |
$$ | X . X O . O O X . |
$$ | . X X X O O O X X |
$$ | X X X . . O O . . |
$$ | . . X O O O O O O |
$$ | X X O O O . . O . |
$$ | O O O . O . . O . |
$$ |-------------------|[/go]

The best sequence seems to be the following:

Click Here To Show Diagram Code

[go]$$W White to play
$$ | X X X X X X X O 6 |
$$ | . . X . O 5 X O O |
$$ | X X X X X O X O 4 |
$$ | X . X O 7 O O X 1 |
$$ | . X X X O O O X X |
$$ | X X X 8 . O O 3 . |
$$ | . 2 X O O O O O O |
$$ | X X O O O . . O . |
$$ | O O O . O . . O . |
$$ |-------------------|[/go]

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?

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



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.



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

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.

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.


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 ?

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.