Book Club: Mathematical Go, Chilling Gets the last point by Elwyn Berlekamp and David Wolfe

shinuito · August 27, 2024, 10:35pm

I think to understand if and when chilling works or not, how it works or if it has limitations in simple endgame it might be nice to come back to Martins examples along with maybe other examples from the book or elsewhere.

So the last time I looked at Martin’s examples, I didn’t really understand what was going on with chilling and with representing it with games, but I’d like to make a second stab at it.

I think firstly, figuring out local sente and gote and move values will help a bit and their connections to CGT.

So I think there’s a couple of ways to compute move values, with swings, with averaging of future outcomes, and then when comparing sente and gote there ends up being some factors of two in some cases depending on counting conventions, which is of course confusing.

I think there’s a few ways of trying to check if a move is a local sente:

If answering doesn’t lose anything, so for example if the expected territory in the position gotten by averaging future positions is the same or smaller than the outcome of just treating the move as sente.
Stanislaw mentioned if the sure gain made by playing is the same size or smaller than the averaged followup.
Stanislaw also put it like if twice the sure gain is the same or less than the continuation computed as a swing.

Eg. 1, G = 0 |-6 || -9

So if White plays it’s 9 points, but if black plays it can be reduced to 6 or 0. If one calculates the expected value of territory with averaging future outcomes, one gets (-9±3)/2=-6 for White.

for (1), the expected is 6 points and treating it as sente guarantees 6 points so that checks out.
for (2), the “sure gain” is 3 points, since black takes away at least 3 points by saving the first stone. The followup averaged is also 3 points, so (2) checks out.
for (3) the “sure gain” is 3 points, and the swing count of the follow up is exactly twice this of 3 points.

So I think you can call connecting the one stone sente for Black.

Eg. 2, G = 0 |-8 || -11

So if White plays it’s 11 points, but if black plays it can be reduced to 8 or 0. If one calculates the expected value of territory with averaging future outcomes, one gets (-11±4)/2=-7.5 for White.

for (1), the expected is 7.5 points and treating it as sente guarantees 8 points so that checks out.
for (2), the “sure gain” is again 3 points. The followup averaged is also 4 points, so (2) checks out.
for (3) the “sure gain” is 3 points, and the swing count of the follow up is 8 points, which is more than twice it.

So I think you can call connecting the one stone sente for Black. It also makes sense, because the follow up is bigger than example 2, while the first move is the same.

Eg. 3, G = 0 | -4 || -9

So if White plays it’s 9 points, but if black plays it can be reduced to 4 or 0. If one calculates the expected value of territory with averaging future outcomes, one gets (-9±2)/2=-5.5 for White.

for (1), the expected is 5.5 points and treating it as sente guarantees only 4 points it doesn’t pass check (1).
for (2), the “sure gain” is 5 points for saving two stones. The followup averaged is only 2 points, so (2) doesn’t check out.
for (3) the “sure gain” is 5 points, and the swing count of the follow up is only 4 points, which isn’t more than twice it.

So I think connecting the two stones is locally gote for Black. Maybe it makes sense, because some of the follow up value has kind of been transferred into the intial move, so the follow up seems smaller.

A bit on chilling:

That’s whenever the game itself isn’t already a number. Numbers don’t get chilled it seems. x_t=x to t>=0 if x is a number.

Also if G_t is arbitrarily close to a number x at some value t, then G_t’=x for all t’>t.

For Martins problems

Game 1

So for the first game, Game 1 (just to distinguish them), with moves A and B, there’s two hot subgames at A = 0 | -8 and at B = 9 || 4| 0 .

Now in terms of trying to understand these, one can calculate their mean and temperatures. The mean of A is m(A)=-4 and t(A)=4, while for B it’s m(B)=11/2 and t(B)= 7/2.

These would agree with the idea of A and B being gote and calculating as in the previous examples. One can also verify this by chilling games.

A_t= -t | -8+t and when t~4 that’s when the left and right values become the same: A₄= -4 | -8+4 = -4 | -4 = -4+ * which is abitrarily close to the number -4.

So this is the mean value m(A)=-4 that the postion cools to, while the temperature is the value of the cooling to reach this number, which is 4, t(A)=4.

The meaning in this case is that the temperature is like an incentive to move, but also, since it’s also the swing from the mean to either of the outcomes, it is also the move value in a Go sense.

For B maybe the cooling is trickier. B = 9 || 4| 0, then B_t= 9-t || {4|0}_t+t.

Now {4|0}_t = {4-t | t} when t<2, and then for t=2 it’s {2 | 2} and afterward it’s just the number 2. So it’s still a hot game and not a number when t<2.

We know from the number translation theorem though that

So that B_t= 9-t || {4 -t | t} +t is

B_t= 9-t || {4 | 2t } for t<=2,

and

B_t= 9-t || 2+t for t >=2,

So that when t=3.5 or 7/2 then 9-3.5 = 5.5 = 2+3.5, so that B_3.5={11/2 | 11/2 } = 11/2 +*.

So this is the mean value m(B)= 11/2, and t(B)=3.5 is the temperature.

So I think during the chilling process the temperature/move value is dropping until the game becomes a number.

In terms of just playing the biggest moves, Black or White should play A since the move value or temperature is t(A)=4 which is bigger than t(B)=3.5

If we want to chill the game, and play the chilled game, work with chilled values etc, we need to tax away a certain number of points. It makes sense then that this corresponds to the temperature of the game, though I’m curious when the temperature is a half integer should you still tax an integer amount? Maybe less, because chilling too much kills the infinitesmals in the position?

So in the above I tax four points to white’s area at A, and 3 to Black’s at B.

Then in the chilled game whoever moves adds a marking. So rather than white getting 8, they get taxed by 5, and black when saving removes a white marking. A => {3 | -3}.

I think, if it’s territory, the white markings cancel out the white territory, but when it’s not, the white markings are like a negative to white.

I’ve also been reading a little https://publications.csail.mit.edu/lcs/pubs/pdf/MIT-LCS-TR-348.pdf where they have a section on Switch games {x|y} with x>y. They mention that you can unbias it by converting it to the form u+{v|-v} where u=(x+y)/2 and v=(x-y)/2. So then it should be -4 +{4|-4}. I guess adding four markings is taxing away the -4, and then chilling the leftover game by 1 is how you end up at {3|-3} above. If you chill by 4 instead of 1 you’d get *.

For B the game with three markings it’s like 5 || 1 | -1, or with four markings 4 || 0 | -2, which it turns out is just the same as subtracting off the number 1. I guess that’s the point in that taxing is just subtracting off a certain amoint of points to begin with, while chilling is the marking per move. Since it’s one marking per move, it’s just chilling by 1.

You could take the game B=9 || 4 | 0, center it by subtracting 11/2, that is 9 || 4 | 0 = 11/2 + 7/2 || -3/2 | -11/2. Then you could chill down the remainder by 7/2 say, so that it’s 7/2 -t || 2+t when t>2 and again I think it cools to * in that case. It’s really just a repetition of the previous things mentioned.

I think my conclusion of looking at this is that chilling doesn’t really help once the games are hot enough and have a high enough temperature?

Game 2

So the only difference is the additon of C = {0| -4}, which has mean m(C)=-2 and temperature t(C)=2.

So how does that change things? I guess the temperature is one thing, the fact there’s a sequence which has a followup is another.

Sorting by temperature t(A)=4, t(B)=3.5, t(C)=2. When B is played by white the temperature drops to 2, which makes the followup to B miai with C. So taking the highest temperature for black would lead to a draw.

On the other hand taking C, leaves the game A+B from before, which I guess is a tie with white moving first. So then B? I’m not sure there’s any magic indicator.

One could also look at the means. m(A)+m(B)+m(C)=-4+5.5±2=-1/2, while m(A)+m(B)=1.5. Maybe there’s indication of score and rounding there?

Another thing https://publications.csail.mit.edu/lcs/pubs/pdf/MIT-LCS-TR-348.pdf mentions is strategies on sums of switch games and a bit about Go at some point.

In fact on page 78 it says:

However, though fractional values are used to estimate Go positions, fractional values never actually appear in Go.

I’m not really sure I’ve gotten anywhere conclusive honestly, but sure look.