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 Post subject: A model to understand komaster and komonster
Post #1 Posted: Mon Jan 03, 2022 2:03 pm 
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Komaster is a term developed by Berlekamp & students. It regards a ko where one side has so many more ko threats that they can win the ko even without losing any points even if they haven't yet ended the ko.

In this post, I solve
a) Direct ko
b) Direct ko + RIGHT enlargement

in the cases

i) Neutral threat environment
ii) RIGHT is komaster (some komonster)

NTE appears in Bill's paper (see next post), and I misunderstood it in this post, so although my calculations are close to the right answer, they are simpler than the correct NTE calculations. I suspect that the examples in this post are simple enough that my calculations are sufficient. However, in general, my method should give results that are close to correct, but slightly off.

Direct ko

Neutral threat environment

A direct ko has 4 positions, A-B-C-D. Say the two players are LEFT and RIGHT. Let the ko swing be K, so the difference in points between winning and losing the ko is K points.
Say LEFT moves left locally when they can and RIGHT moves right. We score for RIGHT. Let A have score 0 and D have score K.

Normally if ko threats are balanced, we can estimate this as

A = 0 , B = K/3, C = 2K/3, D = K

by taking the average. The average can be argued by Jasiek's summary that all moves values (gote/sente etc) use the same unit in miai counting and playing the move with highest gain is such a good default strategy that we can assume as an estimate that a transition from B to C is only worth playing if a transition from B to A by RIGHT is worth playing by LEFT. This forces us to count B as (A+C)/2). The gain of every move here is K/3 in miai counting (or 2K/3 in deiri swing).
(Thermography provides a model to "prove" these numbers, at temperature T=0)

Komaster
However, if RIGHT is ko master, then we can value as if RIGHT has already won the ko in position C. Hence, our new estimate looks more like

A = 0, B = K/2, C = K, D = K.

(This doesn't necessarily mean RIGHT will actually win the ko in position C, but only that such a result is the optimal result for both, so perhaps RIGHT can lose the ko and get a compensation of K elsewhere).

Temperature
Technically we also need to include a temperature by including a tax of T for every move played and a bonus of T/2 for sente. We can think of T as the largest move in the environment of the ko.
NTE
So normally we might have

A = 3T/2, B = K/3+T/2, C=2K-T/2, D = K-3T/2

When K<3T, A>D, so it isn't worth playing in the ko.

Ko master
A = 3T/2, B = K/2+T/2 , C = K-T/2, D = K-3T/2

Ko monster
This means that RIGHT has so many ko threats that they can even keep playing in the environment (respond to every move in the environment that the opponent resorts to playing as a ko threat) until the temperature goes down to t<T.
Now we might have something more like
A = 3T/2, B = K/2+(3T-t)/4 , C = K-t/2, D = K-3t/2
where all the values have shifted in RIGHT's favour.

Hopefully this was a correct understanding.

Research problem
Can you construct a theory of how to draw the thermograph of the sum of two games? Or some principles beyond the obvious?

I have proved that it is impossible given only the information: thermograph of each subgame. However, which aspects go into the combined thermograph?

Now for some more tentative analysis that might be stretching my understanding too far.

Evaluating a multi-stage ko seems a bit like an extension of evaluation of a hyperactive ko. This one where only in the C position, both players have another option. RIGHT may risk moving to E where LEFT has many local ko threats but might gain K+L upon winning the ko, or LEFT might defend at E by moving to say F at the cost of a move instead of moving to B (this removes the possibility of RIGHT winning E entirely).

Of course, RIGHT can't profit from this in a neutral threat environment since LEFT has local threats, but it will change the equilibrium temperature of the ko.

Direct ko + RIGHT enlargement option

NTE

If RIGHT's best play is to end the ko with D, then since D leaves a threat of {0|L-T} behind (if L>T) then we can assume playing L is the largest move. (if L<T, we can forget the possibility of playing E).
If L>T, LEFT must defend and at D, RIGHT expects to gain a move even though they don't get L.
A = 3T/2, B = (K+T)/3 + T/2, C = 2(K+T)/3 - T/2 , D = K+T - 3T/2

If RIGHT's best play is to play one move towards E from C, then we just write E on the right, and enter a different stage of ko where RIGHT has spent a move (like an approach ko) but has hooked on E.
A2 = T/2, B2 = (K+L)/3 - T/2, C2 = 2(K+L)/3 - 3T/2 , E = K+L - 5T/2

Since the transition occurred at state C, we want to know when equilibrium occurs, namely when C=C2 (with a transition)

we have C2-C = 2(L-T)/3 - T = 0 so L=5T/2 is the critical loss above which RIGHT wants to play towards E. We can see (as expected) that this is pretty large. This is because playing normally to D would have gotten profit from the existence of E anyway. Curiously, it is independent of K.

Similarly, if LEFT's best play is to play one move to F from C, we again enter a different stage of ko
A3 = 5T/2, B3 = K/3 - 3T/2, F = C3 = 2K/3 + T/2 , D3 = K - T/2

Equating at C again, we have
C3-C = 2T/3 - T = 0. So T=0. So only when T=0 does LEFT consider to block and wipe out the possibility of E completely. (unless C2 is relevant)
C3-C2 = 2T - 2L/3 = 0. So L=3T. When L>3T, LEFT prefers to block the possibility of E.

Summary
- T=0: LEFT can block and move to F at no cost if RIGHT might win the ko
-L<T: Both sides should treat the threat of E as a bonus of (L/2) added onto the size of the ko (with K->K+L/2). As usual, moves gain K/3. If K_old + L/2 = K_new>3T, the ko is worth fighting. So the ko is a bit bigger with higher temperature. RIGHT's profit from the existence of E is always rho x L/2 (which appears at T=L/2), where rho = 0 in A, 1/3 in B, 2/3 in C, 1 in D. C=2(K+L/2)/3-T/2
-T<L<5T/2: In this case, connecting at D is sente and K_new = K_old+T. RIGHT's profit from the existence of E is rho x T. C = 2(K+T)/3 - T/2
-5T/2<L<3T: RIGHT ignores D and fights the ko directly. C = 2(K+L)/3 - 3T/2
-L>3T: LEFT blocks the possibility of RIGHT enlarging the ko. C = 2K/3 +T/2

It is most interesting that there is a sweet spot at 5T/2<L<3T where fighting over E occurs. This is because it is large enough to be interesting for RIGHT but also LEFT has to pay a penalty of T to prevent it. This is why they say "Don't play go if you fear ko." The point is that, preventing ko variations has a cost, and hence there is a region where playing ko is optimal for both sides. Only if the ko is really big should LEFT spend a move to prevent it.

RIGHT Ko master
A = 3T/2, B = (K+T)/2 + T/2, C = K+T - T/2 , D = K+T - 3T/2
A2 = T/2, B2 = (K+L/2 - T/2, C2 = K+L - 3T/2 , E = K+L - 5T/2
A3 = 5T/2, B3 = K/2 - 3T/2, F = C3 = K + T/2 , D3 = K - T/2

C2=C when L=2T
C3=C always
C3=C2 when L=2T

Summary
We have C = K+T/2 in all cases
-L<2T: RIGHT shouldn't try to enlarge the ko (RIGHT will win the ko and threaten L anyway). Similarly, LEFT should not both stopping RIGHT from enlarging the ko.
-L>2T: RIGHT should try to enlarge the ko. However LEFT should never give RIGHT the chance and should immediately block the ko enlargement by spending a move.

_________________
Give me triangles strong enough and I can measure the universe.

When Venus transits, we can align our clocks to one event. By measuring the angle to flat Earth at two places far apart on Earth, we can compute the distance to Venus and the Sun.


Last edited by dhu163 on Mon Jan 03, 2022 6:32 pm, edited 3 times in total.
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 Post subject: Re: A model to understand komaster and komonster
Post #2 Posted: Mon Jan 03, 2022 4:36 pm 
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I tried to attack approach ko with this method and got a critical point of K=3(3/2)^n T with multiplier 3/2 rather than the golden ratio ~ 1.618 that Bill got.

Hence, I think I haven't quite understood the method completely.

I then thought that maybe NTE means that the first player to take a ko always wins that ko and likewise for any future kos bigger than the first ko. The direct ko is bigger than the approach, so it needs bigger threats. However, rereading Bill's paper, I realise Bill assumed the RIGHT could win the ko by playing first. This makes sense because on further investigation, if RIGHT can't win the ko, then taking is pointless as it just means LEFT moves to A with less moves and hence less tax penalty. However, Bill's method gives a clean way to calculate the size of the threat that RIGHT must ignore in order to win, and that is something I haven't completely got to grips with yet.

For a direct ko, little changes, but A = 0+ 2h - T/2 and D = K - 2h + T/2, where h is the size of the ignored ko threat. The NTE condition is that both h are the same, while komaster sets one of the h to zero. Then both sides are equated so K=4h - T. This varies with T so h = (T+K)/4. If h is larger than this then A>D and the threat must be answered. Also h>=T, since we have a bath environment. Hence we also have K>=3T to be worth playing.

With an approach ko, the threat for the second ko needs to be bigger. This time, relative to E, A = 0 + 2h_2 (RIGHT plays larger threat) - 2h_1 (LEFT plays smaller threat) + T/2 (RIGHT has sente), D = K - 2h_2 (LEFT plays larger threat) + 0 (RIGHT responds to smaller threat) - T/2 (LEFT sente).

These are equal at equilibrium (ko threat size), and also to A2 = 0+2h_1 (RIGHT plays smaller threat) -T/2 (LEFT has sente).

So these equal h_2 = 2h_1 - T/2 and 0=K-3h_2 - T/2. So h_2 = (K-T/2)/3, h_1 = (K+T)/6.

At E, LEFT can play to A2 without RIGHT playing a threat (this adds another difference to my simplistic model, since h1>T), so A2 = 0 + T/2, whereas RIGHT can play to F, ignore a h1 threat, then continue to C = ... etc.

Equilibrium (thermo) means that we can get the result when h_1 = T. This occurs when T = K/5 as required, same as Bill. Note that another transition occurs when T rises to h_2 at T = (2/7)K but this is "hidden" because h_2 never comes into play when T>K/5.

Note that this model isn't perfect since we assume threats are large simple gotes. They could be more complicated with one sided follow-ups but then that is not an NTE anymore. I love this result.
__


For n-move approach ko, if RIGHT is ko master and the ko is in LEFT's hand, they can guarantee winning the ko by spending (n+1) moves on it. Likewise LEFT can guarantee winning the ko by spending 1 move on it. This should mean that there is balance approximately when
K-(n+1)T = T, or K = (n+2)T. This is linear rather than exponential.

If RIGHT is ko monster, they will need to keep the ko going and this may lose points (compared to the miai count) (though not relative to the optimal play in the environment alone), which perhaps should be included. If zero points are lost, then we can write balance when K=(n+2)t where t is the lowest temperature RIGHT can sustain the ko until. If some points are lost, then the critical temperature(s) is when K = (n+2)t + loss sustained from T to t. As long as there is a future point where K>RHS(future) and RHS(now)>RHS(future), then it is worth sustaining the ko until we reach the minimum of RHS.

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When Venus transits, we can align our clocks to one event. By measuring the angle to flat Earth at two places far apart on Earth, we can compute the distance to Venus and the Sun.


Last edited by dhu163 on Wed Jan 05, 2022 1:47 am, edited 1 time in total.
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 Post subject: Re: A model to understand komaster and komonster
Post #3 Posted: Mon Jan 03, 2022 6:26 pm 
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There is a proverb

A three-step (approach) ko is not a ko

This says that a three-step ko is basically already settled in favour of one side. Even if not, the expected value that the disadvantaged side can get (given equal threats) is K/13 out of K (according to Bill's Fibonacci numbers), and when T<K/13, it is worth ending the ko.

Big ko threats tend to be removed (under best play) in late endgame, but still sometimes one side will end up with several more ko threats inside the opponent's position and three-step kos remain something to watch out for.

Overall, we have the consequences
  • A three-step ko normally must be worth a minimum of 6T to worth playing in (komaster).
  • In the average case it needs to worth around 13T. So if in middle endgame T=6, then K=78 is approximately the required swing of a ko that is worth fighting/settling. And in middlegame, if T=14, then K=182 is required, more than half the board.

The corresponding numbers for a 4 step ko are 7T, 21T (and T=6 means K=126, T=14 means K=294).

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 Post subject: Re: A model to understand komaster and komonster
Post #4 Posted: Tue Jan 04, 2022 3:52 pm 
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I present a version of Bill's argument for evaluating a 1 move approach ko in an NTE that has "been reduced to algebra".

We have the direct ko, A-B-C-D and the steps

A2-E-F-C

The approach ko is smaller, and equilibrium occurs when the gain of every move in the approach ko is T, the temperature of the environment. (below this temperature, whichever side plays first will gain from the ko).

So we can set A2 = C - 3T locally. (Bill has something like A2 = 0 + 2h_1 - T/2 which includes the environment gain from the ko threat)

We expect moves in the direct ko to have the same gain, but this requires some assumptions about the environment gain. The critical ko threat level h_2>h_1=T/2.
In Bill's model, we have A = 0 + 2h_2 - 2h_1 + T/2 and D = K - 2h_2 + 0 - T/2.
To get to our reduced model we remove the tax (which has been applied in the form of ko threats).
We now have (relative to A2), A = 2h_2 - 2h_1 + 3T/2 = 2h_2 - 5T/2 = 2(h_2 - T) - T/2 and D = K - 2h_2 + 0 + 7T/2 = K - 2(h_2 - T) + 3T/2.

In order for the second ko to have balance at temperature T, we require D-A = 3T. So
K - 4 h_2 + 12T/2 = 3T or h_2 = (K+6T)/4

And A-A2 = T so 2h_2 = 4T and h_2 = 2T and K - 8T + 3T =0 and K=5T as required.

___

The point is that at equilibrium, we can estimate the local gain of every move as T, so E-A2 = F-E = C-E = C-B = B-A = D-C. In a normal ko, we have D-A = K - 4(h-T) and equilibrium, h=T, so the gain of a move is (D-A)/3 = K/3 as expected. We note that as h increases over T, this essentially reduces the swing of the ko because the ko threat compensates for losing the ko.

When we add in the approach ko, we still have (E-A2) = (D-A2)/4 but since D-A2 is not K, we need to understand how it relates to K.

This time, D-A2 = K - 2h_2 + 0 + 7T/2 - 2h_1 + T/2 = K - 2h_2 - 2h_1 + 4T = K - 2(h_2-T) - 2(h_1-T). Can we account for each of the terms? Well, K is obvious. 2h_2 is the ignored threat at D, 2h_1 is the ignored threat at A2, 4T is the number of steps difference. At equilibrium, we need E-A2=T, so therefore
K=2h_2 + 2h_1.

However, we also know that 3T = D-A = K - 4(h_2-T)-2(h_1-T). Hence, K + 3T = 4h_2 + 2h_1. From this, we obtain h_2 = 3T/2 and by applying h_1 = T, we get K = 5T as required.

Generalising this to more complex situations is just a matter of comparing the bonuses of the form (h-T) at each end state. The NTE model sets the assumption that h for the same ko is equal at both ends of the ko.

For two-step kos, the corresponding formulae now look like (switching h_1 and h_2, sorry)
(1) 3T = D-A = K - 4(h_1 - T) + 2(h_2-T) - 2(h_3-T) = K - 4x + 2y - 2z (in general K - 4x + 2y - 2z + ​2w - 2v -...)
(2) T = A - A2 = 2(h_1-T)-4(h_2-T)+2(h_3-T) = 2x - 4y + 2z (in general 2x - 4y + 2z - 2w + 2v - ...)
(3) T = A2 - A3 = 2(h_2-T) - 4(h_3-T) = 2y - 4z (in general 2y - 4z + 2w - 2v + ...)

We substitute z = 0 in (3) so

T = 2y

Next, (2) gives

T = 2x - 2T so 2x = 3T

and (1) gives

3T = K - 6T + T - 0 so K = 8T

This is the same as Bill's result.

For n-move approach kos, (1)+(2) gives 4T = K - 2x - 2y and summing adjacent equations gives a recurrence
2(x/T) = 2(y/T) + 2(z/T) + 2. So the later values look like 0, 1, 3, 6, 11, 19, 32, 53, always 2 less than a Fibonacci number.

Finally K/T = 2(x/T) + 2(y/T) + 4, which will be a Fibonacci number.
__

Actually, Bill's idea here might be genius since balancing different equilibria at the same temperature would be vital for computing influence functions earlier in the game. I would be willing to sponsor someone a few hundred pounds for progress in this area.

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 Post subject: Re: A model to understand komaster and komonster
Post #5 Posted: Wed Jan 19, 2022 1:11 am 
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dhu163 wrote:
a bonus of T/2 for sente.


"Sente" does not express it clearly. You mean "having the turn" or "moving first".

Quote:
We can think of T as the largest move in the environment of the ko.


If the environment is ko-less.

Quote:
NTE
So normally we might have

A = 3T/2, B = K/3+T/2, C=2K-T/2, D = K-3T/2

When K<3T, A>D, so it isn't worth playing in the ko.

Ko master
A = 3T/2, B = K/2+T/2 , C = K-T/2, D = K-3T/2

Ko monster
This means that RIGHT has so many ko threats that they can even keep playing in the environment (respond to every move in the environment that the opponent resorts to playing as a ko threat) until the temperature goes down to t<T.
Now we might have something more like
A = 3T/2, B = K/2+(3T-t)/4 , C = K-t/2, D = K-3t/2
where all the values have shifted in RIGHT's favour.


Please explain how to calculate these [RIGHT-]counts of A, B, C, D!

Quote:
[hide]Research problem
Can you construct a theory of how to draw the thermograph of the sum of two games? Or some principles beyond the obvious?

I have proved that it is impossible given only the information: thermograph of each subgame.


What is your proof? :)

Quote:
Direct ko + RIGHT enlargement option


What is a player's enlargement option? As a go player, I somewhat understand this concept. However, what is it mathematically?

Quote:
NTE

If RIGHT's best play is to end the ko with D, then since D leaves a threat of {0|L-T} behind (if L>T)


Is this what you mean as a player's (here: RIGHT's) enlargement option?

Previously, you have stated the count K of D. Now, K = ((L - T) + 0) / 2 = (L - T) / 2. Do I understand this correctly?

Quote:
then we can assume playing L is the largest move. (if L<T, we can forget the possibility of playing E).


Is E the position and count after RIGHT moving from the position D to the count L - T?

Quote:
If L>T, LEFT must defend


At which position must LEFT defend? Why?

Quote:
and at D, RIGHT expects to gain a move


In mathematical terms, what is "to gain a move"? Do you mean "move" or [a] "play"?

Quote:
even though they don't get L.


What do you mean?

Quote:
A = 3T/2, B = (K+T)/3 + T/2, C = 2(K+T)/3 - T/2 , D = K+T - 3T/2


How to calculate these counts?

Quote:
If RIGHT's best play is to play one move towards E from C,


I suppose "in 2 net plays, one from C to D, then the second from D to E".

Quote:
then we just write E on the right,


Do you suggest writing A - B - E? Or A - B - E - E? Or A - B - E - E - E? Otherwise, what do you mean by writing the count E on the right?

Quote:
and enter a different stage of ko where RIGHT has spent a move (like an approach ko) but has hooked on E.


What is a "stage of a ko" in your context? What is "spending a move"?

Quote:
A2 = T/2, B2 = (K+L)/3 - T/2, C2 = 2(K+L)/3 - 3T/2 , E = K+L - 5T/2


How and why to calculate like this?

Quote:
Since the transition occurred at state C, we want to know when equilibrium occurs, namely when C=C2 (with a transition)


Is equilibrium a CGT term or your informal description? If it is a term, what is its definition?

What transition?

Quote:
we have C2-C = 2(L-T)/3 - T = 0


C2 - C = 2(K+L)/3 - 3T/2 - (2(K+T)/3 - T/2) = 2K/3 + 2L/3 - 3T/2 - 2K/3 - 2T/3 + T/2 = 2L/3 - 2T/3 - 3T/2 + T/2 = 2(L-T)/3 - T, which you then compare to 0. I see.

Quote:
so L=5T/2


2(L-T)/3 - T = 0 <=>

2(L-T)/3 = T <=>

L - T = 3T/2 <=>

L = 5T/2.

Right.

Quote:
is the critical loss above which RIGHT wants to play towards E.


Please explain what you mean with critical loss and why then he wants to play to E!

Quote:
We can see (as expected) that this is pretty large.


Large in comparison to what?

Quote:
This is because playing normally to D would have gotten profit from the existence of E anyway.


Please explain this in greater detail!

Quote:
Curiously, it is independent of K.


Interesting indeed.

Quote:
Similarly, if LEFT's best play is to play one move to F from C,


What is F? Do you mean if, instead of RIGHT's extension, we have LEFT's extension?

Quote:
we again enter a different stage of ko
A3 = 5T/2, B3 = K/3 - 3T/2, F = C3 = 2K/3 + T/2 , D3 = K - T/2


How to calulate these counts?

Quote:
Equating at C again,


Why?

Quote:
we have C3-C = 2T/3 - T = 0. So T=0.


C3 - C = 2K/3 + T/2 - (2(K+T)/3 - T/2) = 2K/3 + T/2 - 2K/3 - 2T/3 + T/2 = -2T/3 + T so what you calculate is rather C - C3 :) Anyway, you then compare to 0 and find T = 0.

Quote:
So only when T=0 does LEFT consider to block and wipe out the possibility of E completely.


Interesting.

However, I do not understand: When you consider F at all, do you assume the existence of E? I thought you would be studying F instead of E.

Quote:
(unless C2 is relevant)


Please explain.

Quote:
C3-C2 = 2T - 2L/3 = 0.


C3 - C2 = 2K/3 + T/2 - (2(K+L)/3 - 3T/2) = 2K/3 + T/2 - 2K/3 - 2L/3 + 3T/2 = 2T - 2L/3, which you then compare to 0. I see.

Quote:
So L=3T.


2T - 2L/3 = 0 <=>

2T = 2L/3 <=>

T = L/3 <=>

3T = L.

Right.

Quote:
When L>3T, LEFT prefers to block the possibility of E.


Please explain!

Quote:
Summary
- T=0: LEFT can block and move to F at no cost if RIGHT might win the ko


Why?

Quote:
-L<T: Both sides should treat the threat of E as a bonus of (L/2) added onto the size of the ko (with K->K+L/2). As usual, moves gain K/3. If K_old + L/2 = K_new>3T, the ko is worth fighting. So the ko is a bit bigger with higher temperature. RIGHT's profit from the existence of E is always rho x L/2 (which appears at T=L/2), where rho = 0 in A, 1/3 in B, 2/3 in C, 1 in D. C=2(K+L/2)/3-T/2
-T<L<5T/2: In this case, connecting at D is sente and K_new = K_old+T. RIGHT's profit from the existence of E is rho x T. C = 2(K+T)/3 - T/2
-5T/2<L<3T: RIGHT ignores D and fights the ko directly. C = 2(K+L)/3 - 3T/2
-L>3T: LEFT blocks the possibility of RIGHT enlarging the ko. C = 2K/3 +T/2


Please explain each bit of these statements in great detail, thanks!

Quote:
It is most interesting that there is a sweet spot at 5T/2<L<3T where fighting over E occurs. This is because it is large enough to be interesting for RIGHT but also LEFT has to pay a penalty of T to prevent it.


Please explain with more words why this is so in this range!

Why are equalities excluded?

Quote:
Only if the ko is really big should LEFT spend a move to prevent it.


Why is this so in terms of values?

Quote:
RIGHT Ko master
A = 3T/2, B = (K+T)/2 + T/2, C = K+T - T/2 , D = K+T - 3T/2
A2 = T/2, B2 = (K+L/2 - T/2, C2 = K+L - 3T/2 , E = K+L - 5T/2
A3 = 5T/2, B3 = K/2 - 3T/2, F = C3 = K + T/2 , D3 = K - T/2

C2=C when L=2T
C3=C always
C3=C2 when L=2T


Please explain these calculations! How to calculate the values? Why are these the calculations to do? What shall the comparisons express and how are they derived?

Quote:
Summary
We have C = K+T/2 in all cases
-L<2T: RIGHT shouldn't try to enlarge the ko (RIGHT will win the ko and threaten L anyway). Similarly, LEFT should not both stopping RIGHT from enlarging the ko.
-L>2T: RIGHT should try to enlarge the ko. However LEFT should never give RIGHT the chance and should immediately block the ko enlargement by spending a move.


Why?

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 Post subject: Re: A model to understand komaster and komonster
Post #6 Posted: Wed Jan 19, 2022 3:21 pm 
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Thanks for the reply. I have thorough responses in order.

I'm glad that through this process, I came to understand the NTE theory as well as derive the Fibonacci result (which I do believe in as the "correct" solution even if some of the NTE proof might seem slightly handwavy). Don't take my results at the top of this thread too seriously.

  • I like "having the turn".
  • ko-less, sure
  • My counts are just a simplified model, and less well-defined than Bill's thermography. But I hope they give the same result at the equilibrium temperature. I try to measure the expected value of the ko when the ko moves gain less than T (the value of moves in the environment). Then, I just add on a tax to balance the transitions. The ideal is to model the ko as if it merely consisted of 4 states which have definite score. Then it acts just like a normal finite combinatorial game. And then, the decision about whether to play in the ko or the environment reduces to standard theory.
  • I don't remember, but I don't think its hard, just add some depth to variations, and some will be hidden in their own thermographs, but they will interact when added, producing a different result than simpler games that are represented by the same thermograph.
  • I mean that in addition to the ko, if the position is in state D, then this creates an extra {0|L-T} endgame compared to if it were in state A. The difficulty with the ko is that the position is hyperactive if L>T, since even if the position is in state C, RIGHT may be able to try to get the L-T gain before ending the ko. Likewise, LEFT may play to 0 (left in {0|L-T}) in order to prevent this, rather than taking the ko. I realise that to be consistent, I should should have written {T|L-T} which will lead to slightly different results.

    How to write this mathematically? How about (without a tax)
    A = 0, B = {A|C}, C = {B, C_L|D, C_R}, D = K + {0|L} = {D_L|D_R},
    B_L = {A|C_L}, C_L = {B_L|D_L}, D_L = K,
    B_R = {A|C_R}, C_R = {B_R|D_R}, D_R = K+L.
  • No, it is more like K -> K+(L-T)/2, but the exact details need to be worked out.
  • I suppose E = D_R above
  • I mean that if RIGHT wins the ko by moving to D, then if L>T, then a play in {0|L} is the best move. However, now I see, I was wrong and it should be L>2T. I don't know what this implies for the rest. Probably it merely shifts the bound to 2T<L<5T/2
  • I mean that LEFT plays in {0|L}, so RIGHT has sente, giving them an extra T compared to normal. How do you distinguish move/play?
  • I mean LEFT plays in {0|L} to 0, rather than RIGHT playing to L
  • I'm only 80% sure of what I'm doing here. I was trying to relate this to a normal ko. Since LEFT has to play another move in {0|L} if RIGHT wins the ko (at D), then the swing of the ko is effectively K+T
  • No, C to C2 (=C_R), then C2 to E (=D_R)
  • My thinking is clearer now, so see the above with D_R and D_L. I guess D_R has a higher local value than D for RIGHT, so you can think of it as pulling the ko towards the right.
  • "stage" is used in a different context, but I think Charles Matthews used this term or similar in a paragraph on the graph theory of approach kos. Perhaps level is more appropriate? In general, each new game like {0|L} adds a new dimension, but in this simple context, we can see it as adding two new levels to the ko. To move from C to C_R, RIGHT needs to spend a move and likewise LEFT from C to C_L (even though they all take two moves by LEFT to reach A).
  • The K+L is based on the swing of the ko, but a T is subtracted as tax because RIGHT has spent a move from C to C2=C_R
  • This is my attempt to extend Bill's NTE work to more situations. My attempt to simplify is to find the temperature at which the two levels align. At this temperature, there will be a transition between it being worthwhile for RIGHT to play from C to C_R and it not being worthwhile. This is a slightly more general concept than equilbrium (temperature) in thermographs, but I intend for them to be equal when both are well-defined.
  • good
  • good
  • well, below that temperature, RIGHT doesn't want to play towards E (if T=0, then RIGHT prefers C to C2, and no transition occurred before). Hence, after a transition (unless degenerate, but the equation is linear not quadratic or higher, so a repeated root is impossible), RIGHT wants to play towards E and from C to C2.
  • L>=5T/2 is large compared to 2T (environment is like lots of {0|2T}), but I guess it's not that large.
  • I mean that compared to L=0 (i.e. if RIGHT's extension didn't exist), the value of D has already gone up somewhat in RIGHT's favour. This means that even if RIGHT doesn't move to E=D_R (and win L), RIGHT can still get indirect profit via D. Similarly, RIGHT should expend less effort trying to move to E compared to D.
  • yes, I can't explain clearly why. I guess the ko acts as a sort of spring that holds some of the tension that {0|L} creates, and the additional "tension" only depends on L and the structure of the ko, and not on K.
  • Yes, F=C3=C_L
  • see above.
  • see above, to get transition temperature.
  • T=0 is the transition. True, I have a sign error, probably because I was thinking on LEFT's behalf. At T<0, LEFT is interested in moving to F compared to E. For L>>T>0, LEFT is no longer interested.
  • I mean that even though LEFT not prefer C3=C_L to C (without the presence of C2=C_R), LEFT might prefer C3 to C2.
  • Good
  • Good
  • We have a transition. If L is small and positive, LEFT is not interested in moving to C3=C_L. Hence, after transition, LEFT prefers to move to C3.
  • Transition as above. Perhaps T<=0 is appropriate instead.
  • Only if you under the transition point.
  • Equalities excluded for convenience, these problems are a bit like computational geometry (as all optimisation problems are), and I think ignoring degenerate cases makes for cleaner mathematics. In Go, the problems are simple enough that equality normally simply means you have more than one optimal strategy, but I'm not completely confident about what happens in general.
  • I meant that only if L is really big, i.e. L>3T, should LEFT play a move to prevent it. However, if L really is that big, in Go language, we often refer to L as part of the ko anyway.
  • assuming my original calculations were correct, see above.
  • ditto

So in responding, I have gained some more confidence, noticed a mistake ({T|L-T}), and realised that the ko merely acts as a spring that stretches due to the presence of L. What more magic is hidden in Go?

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Post #7 Posted: Wed Jan 19, 2022 10:49 pm 
Judan

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I need to think more about it:)

You mention Bill's paper(s) in the book Computers and Games: Third International Conference, CG 2002 - is this paper available as PDF somewhere or must one buy the book? He also wrote about NTE / Fibonacci in his paper in the 4th ICOB (International Conference on Baduk), which is / was also available online since some years after the conference. Are there more NTE papers of his elsewhere?

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Post #8 Posted: Thu Jan 20, 2022 10:28 pm 
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I found that link on SL (as usual). That's all I know.

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Post #9 Posted: Sat Jan 22, 2022 6:54 pm 
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RobertJasiek wrote:
I need to think more about it:)
You mention Bill's paper(s) in the book Computers and Games: Third International Conference, CG 2002 - is this paper available as PDF somewhere or must one buy the book?


Evaluating Kos in a Neutral Threat Environment: Preliminary Results


This post by dany was liked by: dhu163
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Post #10 Posted: Sun Jan 23, 2022 12:18 am 
Judan

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Thank you!

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