Thermography

Gérard TAILLE · Post by **Gérard TAILLE** » Fri Jun 18, 2021 9:51 am

dhu163 wrote:Thanks for your explanations. They all make sense, especially the variation with on the monkey jump. That ko does seem to be critical, so my "proof" is flawed for missing it out.

I meant for "reverse" to be the same as the SL definition.

Instead of "if not" can black with the move win [the difference game], I tried to prove "can white on the previous move draw with a local response" which is equivalent IIUC. Though the colors in my diagram are switched relative to SL.

As I mentioned earlier the problem with the position studied and the different moves is that, in many cases, a ko appears in the sequence. As a consequence we cannot use the result of CGT. Because reversal is only defined where they are no kos we cannot use this term as it is defined. For the same reason we cannot also use the terme "dominate".
That is the reason why I formulate my question by avoiding using such terms.

Thank you for having tried to answer my question. Unfortunately it appears that the question remains open. Not easy is'it?

Gérard TAILLE · Post by **Gérard TAILLE** » Tue Jun 29, 2021 8:19 am

Click Here To Show Diagram Code: [go]$$B Diag 1 $$ --------------------- $$ | X O O . O . . . . . $$ | X X X X O . . . . . $$ | . . . . . . . . . . $$ | . . . . . . . . . . $$ | . . . . . . . . . .[/go]

Click Here To Show Diagram Code: [go]$$B Diag 2 $$ --------------------- $$ | . O O . O . . . . . $$ | X X X X O . . . . . $$ | . . . . . . . . . . $$ | . . . . . . . . . . $$ | . . . . . . . . . .[/go]

The positions in diag1 and diag2 correspond to the trees {2|-2} and {2½|-2½} (after normalization around 0)

But what happen with chinese rule?
The positions in diag1 and diag2 correspond to the trees {3|-3} and {3½||-2½|-4½}.
My question is the following. Can we build a position corresponding (in chinese rule) to {3½|-3½} after normalization ?

dhu163 · Post by **dhu163** » Fri Jul 09, 2021 11:19 am

Gérard TAILLE · Post by **Gérard TAILLE** » Fri Jul 09, 2021 1:09 pm

dhu163 wrote:
$$
$$ ---------------------
$$ | . a O X . . . . . .
$$ | X . O X . . . . . .
$$ | O O O X . . . . . .
$$ | . O X X . . . . . .
$$ | O O X . . . . . . .
$$ | X X X . . . . . . .
Click Here To Show Diagram Code
[go]$$ $$ --------------------- $$ | . a O X . . . . . . $$ | X . O X . . . . . . $$ | O O O X . . . . . . $$ | . O X X . . . . . . $$ | O O X . . . . . . . $$ | X X X . . . . . . .[/go]
A move at a.

The one eye each seki is required for the parity of (3 1/2 - (-3 1/2)=7) to be odd(unless there are relevant intricacies to Chinese rules I am not aware of)

As far as I know you can't make a smaller eye than this on a Cartesian grid, so I'm guessing 7 is the smallest odd number possible.

Click Here To Show Diagram Code: [go]$$B $$ --------------------- $$ | . . O X . . . . . . $$ | X . O X . . . . . . $$ | O O O X . . . . . . $$ | . O X X . . . . . . $$ | O O X . . . . . . . $$ | X X X . . . . . . .[/go]

This position does not correspond to the pure {3½,-3½} because after:

Click Here To Show Diagram Code: [go]$$B $$ --------------------- $$ | . 1 O X . . . . . . $$ | X b O X . . . . . . $$ | O O O X . . . . . . $$ | . O X X . . . . . . $$ | O O X . . . . . . . $$ | X X X . . . . . . .[/go]

There is still room for a following move like for example a black move at "b" as a good ko threat. Such following move cannot exist with a pure {3½,-3½} gote point.

dhu163 · Post by **dhu163** » Fri Jul 09, 2021 5:27 pm

Click Here To Show Diagram Code: [go]$$B $$ --------------------- $$ | . 1 O X . . . . . . $$ | X b O X . . . . . . $$ | O O O X . . . . . . $$ | c O X X . . . . . . $$ | O O X . . . . . . . $$ | X X X . . . . . . .[/go]

I don't believe in the existence of such a pure {3½|-3½} then since I think seki is required.
Note that even white c is sometimes a (super)ko threat. So I think worrying about ko is overkill most of the time.

To clarify the parity point, in Chinese rules, in a fixed area of say n intersections, when a position is settled, If there are no seki neutral points, m are controlled by black and n-m by white. Using stone counting the score for black is m -(n-m) =2m-n +C, where C is a constant to "normalise" the score.

As m varies, this only changes by multiples of 2.

For example in your {3 1/2 ||- 2 1/2 |-4 1/2} example, we have 3 1/2 -(-2 1/2)= 6 which is even.

Gérard TAILLE · Post by **Gérard TAILLE** » Sat Jul 10, 2021 10:22 am

dhu163 wrote: The one eye each seki is required for the parity of (3 1/2 - (-3 1/2)=7) to be odd(unless there are relevant intricacies to Chinese rules I am not aware of)

As far as I know you can't make a smaller eye than this on a Cartesian grid, so I'm guessing 7 is the smallest odd number possible.

Click Here To Show Diagram Code: [go]$$B $$ --------------------- $$ | . O O a . X O . . . $$ | X X . X X X O . . . $$ | O X X X O O O . . . $$ | O O O O O . . . . . $$ | . . . . . . . . . . $$ | . . . . . . . . . .[/go]

I do not understand why parity requires one eye each and seki. What about the example above and a move at "a"?

dhu163 · Post by **dhu163** » Sat Jul 10, 2021 1:03 pm

Because sekis between groups with no eyes have 2 neutral points, an even number, which maintains the parity property.
A seki between two groups of one eye each has 1 neutral point, which allows for odd number differences between scores.
Such simple sekis tend to require each group having exactly 2 liberties after all neutral points are filled, so when groups have one eye, the eye offers one of the liberties and they only need one more neutral point as a mutual liberty.
(Incidentally, this is why you can't have a seki between a group with one eye and a group without an eye. However, you can chain sekis together with
one eye - no eye - no eye - ... - no eye - one eye, where adjacent groups share one liberty.)
Without seki, there are no neutral points, and zero is an even number.

There are more complicated types of seki and I don't know how many neutral points they have.

In your example, I'll just stone count the black stones and internal region.
If black plays at a, they control all 15 points
If white plays at a, white controls 3 stones, black 9. This seems like 6 points with 3 neutral points and 15-6=9 which is odd, contradicting what I said above. However, after a white move at a, the position is still not settled under Chinese rules, and during dame filling, one side will play on one of the neutral points. If white plays it, they control 4 stones, leading to a score of 9-4=5. If black plays it, the score is 10-3=7.

Parity is maintained since 15,5,7 have the same parity (they are all odd).

i.e. I claim your example shows {15||7|5}

Gérard TAILLE · Post by **Gérard TAILLE** » Sun Jul 11, 2021 2:49 am

Click Here To Show Diagram Code: [go]$$B $$ --------------------- $$ | . O O a . X O . . . $$ | X X . X X X O . . . $$ | O X X X O O O . . . $$ | O O O O O . . . . . $$ | . . . . . . . . . . $$ | . . . . . . . . . .[/go]

dhu163 wrote: In your example, I'll just stone count the black stones and internal region.
If black plays at a, they control all 15 points
If white plays at a, white controls 3 stones, black 9. This seems like 6 points with 3 neutral points and 15-6=9 which is odd, contradicting what I said above. However, after a white move at a, the position is still not settled under Chinese rules, and during dame filling, one side will play on one of the neutral points. If white plays it, they control 4 stones, leading to a score of 9-4=5. If black plays it, the score is 10-3=7.

Parity is maintained since 15,5,7 have the same parity (they are all odd).

i.e. I claim your example shows {15||7|5}

I completly agree with you calculation above. But what is my conclusion? The miai value change the parity because when it is white to play the result would be either 7 or 5 that means 6 on average. That is the meaning of my example : you can reach the miai value 4½ (which correspond to 3½ in japonese rule) with a seki without eyes.

Click Here To Show Diagram Code: [go]$$B $$ --------------------- $$ | . O O O a O . . . . . $$ | X X X X X O . . . .. $$ | . . . . . . . . . . $$ | . . . . . . . . . . $$ | . . . . . . . . . .[/go]

Isn't is exactly the same result in the above diagram with no seki, where the game looks like {+5||-3|-5} for a miai value 4½ ?

dhu163 wrote:Because sekis between groups with no eyes have 2 neutral points, an even number, which maintains the parity property.

In practice, no doubt that most of the seki without eyes have 2 neutral points but here we are in theoritical context and sekis may have more than 2 neutral points as in the following famous example with 4 neutral points:

Click Here To Show Diagram Code: [go]$$B $$ --------------------- $$ | X X O O X X . . . . $$ | X O . . O X . . . . $$ | X O . . O X . . . . $$ | X O X X O X . . . . $$ | X O O O O X . . . . $$ | X X X X X X . . . . $$ | . . . . . . . . . .[/go]

You may argue that 4 neutral points is still an even number of neutral points but with more complexe example you can build a seki without eyes, and 3 neutral points.
BTW what is for you the definition of an eye? Maybe a misunderstanding can exist here.

dhu163 · Post by **dhu163** » Sun Aug 01, 2021 5:59 am

By simple sekis, perhaps I meant sekis where
- each neutral point is adjacent to one white chain and one black chain involved in the seki
and
- the adjacency graph of seki chains (where chains are adjacent if they share at least one neutral point) is a tree (i.e. has no cycles).

Restricting to simple sekis, I believe an eye is required for there to be an odd number of neutral points

Definition of eye is quite awkward. The following is not quite adequate for all general examples (such as two-headed dragon), and the recursion might break the definition, but is perhaps closest to how I think:
- connected region enclosed by a chain, where assuming that chain is immortal, opp can't make an eye inside the region.

I suppose that legendary example with 4 neutral points (that "Lee Changho spent an hour on and couldn't solve" was it?) is an example where W has less than one eye but not zero eyes by the above definition. W to play can make eyespace by connecting, but it is dead shape. B to play can cut, but the cut isn't connected to B's other stones, meaning W gets two eyes.

you can build a seki without eyes, and 3 neutral points.

I would be curious to see an example.

Gérard TAILLE · Post by **Gérard TAILLE** » Sun Aug 01, 2021 10:51 am

dhu163 wrote:By simple sekis, perhaps I meant sekis where
- each neutral point is adjacent to one white chain and one black chain involved in the seki
and
- the adjacency graph of seki chains (where chains are adjacent if they share at least one neutral point) is a tree (i.e. has no cycles).

Restricting to simple sekis, I believe an eye is required for there to be an odd number of neutral points

Definition of eye is quite awkward. The following is not quite adequate for all general examples (such as two-headed dragon), and the recursion might break the definition, but is perhaps closest to how I think:
- connected region enclosed by a chain, where assuming that chain is immortal, opp can't make an eye inside the region.

I suppose that legendary example with 4 neutral points (that "Lee Changho spent an hour on and couldn't solve" was it?) is an example where W has less than one eye but not zero eyes by the above definition. W to play can make eyespace by connecting, but it is dead shape. B to play can cut, but the cut isn't connected to B's other stones, meaning W gets two eyes.

you can build a seki without eyes, and 3 neutral points.
I would be curious to see an example.

dhu163, here is a simple example of seki without eyes, and 3 neutral points:

Click Here To Show Diagram Code: [go]$$B $$ ------------------------- $$ | . O . X O O O . X O . . $$ | X O X X X X X X X O . . $$ | X O X O O O O O O O . . $$ | X O X O . . . . . . . . $$ | O O X O . . . . . . . . $$ | X X X O . . . . . . . . $$ | O O O O . . . . . . . . $$ | . . . . . . . . . . . . $$ | . . . . . . . . . . . .[/go]

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