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 Post subject: a slightly silly doubled ko #1 Posted: Tue Nov 02, 2021 11:21 am
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`[go]\$\$ Black to play\$\$ --------------\$\$ | X a b c O .\$\$ | X X O . O .\$\$ | X X O O O .\$\$ | X X . . . . \$\$ | . . . . . .[/go]`

I have noticed that this endgame position is evaluated as 1 1/3 of a point for white.

If white plays b, they get 2 points.
If black plays b, then plays c, white has 0.
If black plays b, then white plays c, white has 1 1/3. At this point only a simple ko remains.

Hence the miai count of the original position is (2+(0+ 1 1/3)/2)/2=(2+ 2/3)/2=1 1/3

So the gain of every move in this position is 2/3.

Alternatively, using iterative deiri counting, the count is (2+0)/2 + (1 1/3 - 0)/4 = 1 1/3.
and the gain is (2-0)/2 + (0 - 1 1/3)/4 = 2/3

So here is the question:

1. Find all endgame positions where the gain of every move below a certain depth is the same.

One example is a simple ko, where the gain of every move is 1/3. In a simple gote, both possible moves gain the same. However, are there more complicated endgames, for example where the gain of every move is 4/3?

`[go]\$\$ Simple ko\$\$ X X O O \$\$ X . X O \$\$ X X O O[/go]`

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 Post subject: Re: a slightly silly doubled ko #2 Posted: Tue Nov 02, 2021 1:26 pm
 Judan

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In each star corridor of arbitrary length, each move has the value 1.

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 Post subject: Re: a slightly silly doubled ko #3 Posted: Wed Nov 03, 2021 4:07 am
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Yes, and by adding repeated dead stones, one can create a STAR corridor for any positive odd integer.

For example for 3, we have

`[go]\$\$ 3 STAR\$\$ --------------\$\$ | X X X X X X X X X O\$\$ | X O O O . O . O . O\$\$ | X X X X X X X X X O[/go]`

In theory we could do something for even integers, but that would require a branching tree, which doesn't really fit on a Cartesian Go board.

4/3 attempts

I tried to do some calculations to construct a position where the first few moves were worth 4/3. However, it didn't work out.

`[go]\$\$\$\$ | O O O . . .\$\$ | O X O . . .\$\$ | O X O . . .\$\$ | X a O O . .\$\$ | X . X O . .\$\$ | X X b O O O\$\$ | . O X . X O\$\$ | . O . O O O\$\$ | . O O O . .[/go]`

Let a,b represent the sizes of those moves (only counting W's solid captures).

This has been set up so that if white spends two moves, white gets a+b+1 (with the half dead ko stone).
If white plays b (b>a), then black plays a, white gets b+1/3
If black plays b, then white plays a, white gets a
If black plays both b and a, white gets -1

Hence, the position if black plays first has a count of (a-1)/2
the position if white plays first has a count of (a+b+1+b+1/3)/2 = a/2 + b + 2/3
The overall count is the average which is (a+b)/2 + 1/3 - 1/4 = (a+b)/2 + 1/12
We require this to be the same as b+1/3

So a/2 = b/2 + 1/4 or b = a + 1/2

Also, since we want the gain to be 4/3, we need (a+b+1)-(b+1/3) = 2* (4/3)
So a = 6/3 = 2
And b = 3/2

Unfortunately, we have contradicted the assumption that b>a.

Corridor constructions

However, at least I did manage to figure out how to construct an endgame of gain 5/2. This is b above.
Using iterative deiri counting, the technique was to gain 2 + 1/2 = 4/2 + 2/4
That is, white playing b wins them 4 points over if black plays and white responds. Next, white's response wins them 2 points over if black played there again.

This is a standard technique. Two corridors are required because if there is just one corridor, a solid entrance move compared to the defender blocking gains an odd number of points (double the number of dead stones it connects to + 1 for the less defence needed) (under territory scoring. If area scoring, then it is always even since it is double the number of intersections that have changed hands, with an extra +1 for the move the invader plays). The subsequent gains down a corridor are positive.

In addition, the sente-gote assumption implies that consecutive gains are monotone decreasing.

Hence the possible gains from a width one corridor (with no liberty issues) is the range of values the sum
S = a_0 + a_1/2 + a_2/4 + ... + a_n/2^n can take.

With the conditions:
- a_0 is an odd positive integer,
- a_i>0, and
- a_n <= a_(n-1) + a_n/2 <= a_(n-2) + a_(n-1)/2 + a_n/4 <= ... <= S

The last inequality means a_(n-1) >= a_n/2, a_(n-2) >= a_(n-1)/2 + a_n/4, etc

For example a valid sequence is 3, 5, 1 because 5>= 1/2 and 3 >= 5/2 + 1/4. In fact 3, 5, 1, 1, 1, ... fits.
It is straightforward to prove that a geometric increasing sequence of the form 1, r, r^2, ... with r>1 does not fit the inequality, whereas r<=1 does.

In particular, S<= a_0 + a_0 = 2* a_0. It is fairly clear that S can take any value in the range [a_0, 2*a_0] if we vary a_i over the reals.

Hence, a corridor can have a gain of [1,2] and [3,infinity]. Unfortunately, we can't seem to produce (2,3) with a single corridor. This is why we need an extra branch to the corridor to create 5/2 as above.

Overall, my conclusion is that I have failed at creating 4/3, and I still don't know if it is possible.

Back to n 2/3

Here is a model.

`[go]\$\$\$\$ | X O X . . . .\$\$ | X O X . . . .\$\$ | X a O O O O O\$\$ | X . X b X . O\$\$ | X X O O O O O[/go]`

If white plays first, they get b
If black plays first (a>b) at a, then again at b, white gets -a
If black plays first at a then white at b, white gets b-a+1/3

To get the required situation, we need 2*(b-(b-a+1/3)) = b + 1/3
=> 2a - 2/3 = b + 1/3
=> 2a = b+1

This is not consistent with a>b unless a<1.

The gain is b-(b-a+1/3)=a-1/3.
We have already found a situation with 2/3 gain. In general, to get n 2/3 gain, we need a = n+1
So b = (n+1)/2 -1 = (n-1)/2.

Conclusion
The doubled ko was probably a special case. I didn't solve my original question, but did find something out about corridors.

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 Post subject: Re: a slightly silly doubled ko #4 Posted: Wed Nov 03, 2021 6:42 am
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I have done some calculations on another position

`[go]\$\$\$\$ | . O . . . . .\$\$ | b O . . . . .\$\$ | O X . . . . .\$\$ | O X . . . . .\$\$ | O X X X . . .\$\$ | a O O X . . . \$\$ | . X X X . . . \$\$ | X O O O O . . \$\$ | c X X . O . . \$\$ | O O O O O . .[/go]`

Assuming a is the best first move, then b then c.

The gain on the first move is a/2 + b/4 + 1/4
if black plays first, the gain of the follow up is b/2 + 1/3
if white plays first, the gain of the follow up is c/2
if black plays first, then black, the gain of the next move is c/2 + 1/6, and if white, the gain is c/2 + 5/6

In order to make these gains decreasing but as close as possible, I have chosen a = 4, b = 6, c = 5.

So the gains are
4 3/4
3 1/3
2 1/2
2 2/3, 3 1/3

Some things to note:

- If a,b,c is not the correct order of moves, the sente-gote assumption is likely to break. Is it possible for two sequences of moves to both satisfy the sente-gote assumption and yet still be optimal play? (I'm guessing yes, but more details)
That is to say that if a move a is sente, forcing b, then if b was playable before a (e.g. an independent endgame), then maybe you should have played b instead of a. If they are dependent, then it is not clear which is best. Often it is a capturing race and both moves are sente. However knowing which to play may depend on the environment and this is when a probing move is useful before choosing.

- Remember that a move's main impact (especially in endgame problems) is on the intersections around it. A move such as a not only captures stones but also sets up b, and reduces the impact of white playing c. Similarly, white playing b to save those stones means that black's stone at a is vulnerable and not connected, leading to further gains if white gets to play c.

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 Post subject: Re: a slightly silly doubled ko #5 Posted: Sat Jan 01, 2022 8:16 am
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I realise this is closely related to Bill's ambiguous kos

https://senseis.xmp.net/?AmbiguousKo gives an example with 1 2/3 gain.

Open problem: Is it possible to construct an ambiguous position with move value (2n+1)+ 1/3?

Also I see that Bill's sente kos have a high intersection with Jasiek's external kos though the context for their usage is fairly different.
Intersection: when a ko can be ended in one move by capturing an opponent chain rather than connecting the ko, i.e. a neighbouring opponent chain only has one liberty
Difference: a sente ko may have a large follow up but can't capture the opponent in just one move. Plausibly, an external ko includes a case where the capture is too small so it isn't sente ko, but I can't think of any example on any board graph.

Again, if my understanding is correct. I would like to put Bill's "hyperactive ko/position" in the context of Jasiek's external kos + graph theory.

Def [hyperactive ko]: a ko that is not an external ko, but nonetheless if one side connects the ko, there is a follow up (larger than the ko) that attacks a neighbouring opponent chain (or otherwise uses its weakness). i.e. If that side is komonster, they don't have to connect the ko and may directly attack the opponent chain.

edit: I can already see my understanding isn't quite right as Bill says approach kos are hyperactive.

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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.

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