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

For the above, note that one also needs to allow the game to end with a fractional score.

(corridors and similar things:
One More Number at Sensei's Library is a fairly complicated example.)

You mean in chilled go? What do you mean by “end”?

If the game is {0|1} = 1/2, it doesn’t have to end there, Black and White both have moves available.

Yes.

" ​ What do you mean by “end”? ​ "

stop, finish, halt?
Alternatively, when for ​ ​ ​ " ​ If the score is zero, whoever played last wins ​ "?

Your last sentence may be a good point: ​ I’ve never seen it presented
that way, but I don’t see any reason why that shouldn’t also work.

I still don’t get it, but I don’t get a lot of things… Why would you have to add new rules about stopping playing with a fractional score? I was imagining you would just stop playing like in normal go but before filling dame.

Would you want to stop before invading the space in that example you linked?

This part I’m not sure is simple is it? As in you’re saying G>0 → G>=1?

One can imagine a game being confused with all kinds of numbers between zero and 1, to potentially make it false, but whether it happens in Go seems tricky.

{3|1}>0 but {3|1} is not >= 1 because it’s confused with 1. But maybe it can’t exist in Go.

I’m not sure one would have to add such rules: ​ ​ ​ That’s just what I
recall seeing for showing almost-equivalence. ​ (eg., excluding ko)

If the space gets scores as its Number - rather than as 0 points
because neither player has anything surrounded - then one would
want to stop before invading the space in that example I linked
because of the 1-point tax: ​ Further moves would be worse than passing.

Right so if we have e.g. a very small half point corridor, which is {1|0}, if it was chilled by 1 point (anything beyond 1/2 really), then it turns into 1/2 the number. Though if there really was a one-point tax to move, then white wouldn’t even want to move and it’s a guaranteed point for black always? Surely you’d only want to move when you can make at least one point?

You don’t get to not move, passing is not a thing in “normal games” and in my interpretation of chilled go. The game ends when one player has no moves.

Normally in CGT there’s no “score” at all, you just keep playing. So if White plays to, say, a 5, Black would play one more time to 4, and the game would then be over.

But I don’t think there’s any practical difference if you just stop at 5.

I think it would be fine to stop at 1/2 too, as you suggest, but I don’t see why you would have to do it.

(As I mentioned, I also don’t:
However, doing so at least least tends to reduce the depths of the summands.)

So maybe this is worth discussing. You talk about taxing multiple points but the whole topic of the book is specifically chilling, cooling by one point. And even more general cooling would apply equally to both sides. You wouldn’t do different numbers of points for the two players.

Right, it’s intended for games where each part is worth about a point.

This would be really interesting to me since I don’t understand any other estimation techniques. Maybe we could pick a couple of positions from the book and analyze them with traditional methods?

A post was merged into an existing topic: Go videos

Well the initial taxing and marking is just shifting the score I think. As in instead of looking at G look at G+1 etc.

I suppose if a game has a half integer mean value, there’s no way to centre it better around zero.

But essentially I was just exploring what happens when you tax away a certain number of points from the positions. It’s more re-centreing the values of each sub part around zero. So if the mean/middle value of a position is 4 points or black, you can tax the position four points to shift it to a mean value of zero.

Different from chilling but used when setting up the chilled game in the book.

Sure I can have a look at some of the example positions. I guess the reason traditional methods stop working is because they’ll all say something like this move is a fraction or a one point move with no way to distinguish between them, while playing the chilled game exposes differences in the form of infinitesimals.

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Previous posts:

So figure 3.1 has a few games, that they eventually say they combine the four of them into one game.

I think 3.1(a) is just figure 2.7

Maybe we can look at (d)? I’ve added in some letters to make it easier to discuss.

So I did try to introduce a few ideas before on how counting or estimating should work.

So there is multiple ways of estimating scores and doing counting. I think one older way was called deiri counting or 出入り（でいり）Deiri Counting at Sensei's Library, which is the term for a kind of swing counting. Google translates it to something like coming and going, entering and exiting. but senseis says it’s more like an old term for income and expenditure.

Anyway I guess the idea is that for gote moves, a move that both makes you points and stops the opponent making points should be bigger than a similar size move that only does one thing. So the swing is the difference between one side playing and the other. I think it’s good enough then for all those kinds of moves to just play them in order from biggest to smallest.

For (B) that means if White plays it’s one point, while if Black plays it’s zero, hence it’s worth 1 point.

For (E) it’s the difference of 2 points to 0, so it’d be 2 points.

I think problematically is that it’s hard to calculate a swing when there’s multiple followups like in (A) etc. Or if a move is sente, comparing it’s size to a gote move. I think you can literally compare them by playing a game with the moves you want to compare on the board and checking the results, though I think sometimes you need to double some positions, as in compare two copies of a move to two of another (something to do with the players needing to spend the same number of moves in each variation - I was watching Stanislaw explain that). I also think in a technical CGT sense, some of these gote moves of the form {x|y} with x>y actually turn into integers/rationals when you double them. That is {x|y}+{x|y}=x+y.

Anyway that’s a sidetrack. You can also count moves with followups by averaging future values as I mentioned in a few previous posts on corridors and other positions. I think this is what is meant by Miai counting at Sensei's Library, or I’ve seen it called Modern or Absolute counting. I guess the idea is that when you average gote moves by their future outcomes it’s putting them on the same scale as sente moves. Typically with swing counting there’s an idea that a sente move or reverse sente move is worth the same as a gote move of twice the size. So a 3 point reverse sente is the same as a 6 point gote. Then I guess in another sense, the miai counting article is saying something like you’re accounting for the fact that in different variations, players might spend more moves than their opponent to make a gain. I guess in the extreme examples of multiple move approach kos, where you have to spend moves to make it a direct ko, you want to account for that somehow, to calculate a true points per move value.

(B) in that case is the smallest corridor, and rather than being 1 as the swing value, the expected territory there is (1+0)/2, the average of the two scored outcomes. Then the move value is the difference from the expected value |1-1/2|=|0-1/2|=1/2.

(C) and (D) are counted the same, but white can play to move to a expected territory of 1/2 or black can take 2 points, so it’s (2+0.5)/2=1.25, and the move value is 3/4.

I think (A), (F) and (G), I think there’s some notion of sente involved. As you do the move calculations, it turns out that say for (F) and (G), the expected territory of the position is the same as if the players just answer as if it was sente. That is they guarantee that amount of territory by answering. For (F) if Black is prepared to answer White they get a minimum of 2 points, while if you do out the averaging (1+3)/2=2 also. Then for (G) it’s similar but 3 instead (4+2)/2=3. So in those cases, I think you’re supposed to take the expected territory for the sente part of the tree as the value of the settled position, that is 2 and 3 respectively for (F) and (G), and the move value is the difference to the reverse sente moves, which is both 1 point more of territory.

Then (A) lastly, I think eventually playing F9 to guarantee 2 points is similar to saving/keeping captured one stone, so blocking it will eventually be sente, and so (A) is like (G), expect 3 points and the move is worth 1 point.

So to summarize that:

## Expected territories and territories

Black:12.5
White:12.5

if you add up all the fractional values and integers.

and then

## Move values

they are all quite close to one point, 1, 0.75, 0.5 point moves.

Then the chilling by 1 point, I suppose exposes infinitesmals, or probably in the gote cases of the corridors they’ve already been made cold by chilling more than their temperature (=move value).

## Chilled game

So to play the chilled game, I think we’re supposed to tax away a certain of points from each position. I think really what this means is that territories are worth 0, and potential territories are centered around zero.

What I mean is, if you took (E) from before, technically it’s {2|0}, because either black can capture for 2 points or white saves the one stone for 0 points to black. If you “tax” away one point, that is add a black marking, then rather than playing H={2|0}, you instead play H-1={1|-1}, which is centered around zero instead.

Maybe it’s confusing to use the word “tax”, because it’s being reserved for chilling? I don’t know what else to call adding the markings though, since you are kind of subtracting or shifting the scores in the positions.

Anyway, adding 12 markings of each color shifts the overall game (G let’s say), to G-12+12=G, so nothing has changed in the overall game, but each position is now better centered around 0.

Then playing the chilled game G1={L1-1|R1+1}, I guess that’s more properly being described as a tax, where when you play a move it costs you one point also, recorded in the form of a marking added or removed from the board.

In any case, the calculations should be reasonably similar to a previous post.

(B) {-1|0}=-1/2, if white plays the mark (tax) balances the one point gain. If Black plays they actually add a black mark to non territory, which is like a negative.

(C)=(D), {0|1/2}=1/4 for a similar reason to the (B) case, but just one step removed.

(E) {0|0}=* because of the markings/tax.

(F) {0|*}=↑

(G) {↓|0}=⇓*=↓+↓+*.

I think (A) isn’t like the examples from previous posts, so I’ll leave it as a ? in the diagram below and compute it separate.

(A) is zero if black plays, but white has a few moves with followups. So it’ll be kind of nested zeros, as each time black plays it’ll be 0, and at the very end if white gets F9 it’s also 0.

So (A) = 0 ||| 0 || 0 | 0 or 0 || 0 | * or 0 | ↑ = ⇑* = ↑ + ↑ + *.

In summary if white plays 3 moves here it removes the three black marks and the position is 0. If black plays at any stage, the territory they get equals the marks, and so it’s also 0. That makes it exactly opposite to (G) for example. Which is kind of the second time I’ve mentioned that

The whole game is then

G = ⇑* - 1/2 + 1/4 +1/4 +* + ↑ + ⇓* = * + ↑ = ↑*, which I think is a first player win.

How does that seem for comparing various counting methods and CGT games?

I think the traditional methods, kind of say to play the moves in order of size, biggest to smallest, except maybe in the case of Tedomari at Sensei's Library, but I guess it doesn’t really distinguish very well between moves that have the same value like all the 1 point moves.

While in the above with CGT, generally you’d want to turn ↑ into * as White so F? Then as black you want the position to be worth ↑ so you should play *, which is E?

I guess the traditional counting might say A and G are miai, and maybe B, C, D have some kind of miai-ness, so it could be E or F, but I guess you have to read it all out?

Anything in particular to flesh out a bit more?

Due to how this turns out, I imagine the example you were looking at
was intended as a complex puzzle of taking all details into account,
rather than just as practice with infinitesimals.

There are the subtleties that

the upper-right is a-priori not immortal, since if White gets all of
J5,J6,G5,D9,E9,F9 , ​ then G9 becomes vital for life/death of the upper-right
and
the loer-right is a-priori not immortal, since if Black gets all of F4,F3,F2 then once the
lower-right runs out of liberties, blackJ2 would make an approach ko for the lower-right

.

regarding the upper-right:

(a)
If the upper-right was immortal, then E+F would be UP*, so White’s best option
there would be settling that sum. ​ whiteJ5 does so, by leaving G5 and J6 miai,
in which case Black gets one of those. ​ White playing G5 instead is no better for White,
since Black can treat whiteJ5 as sente, in which case Black gets one of J5,J6.

(b)
If the upper-right was immortal, then every move in the A,E,F components would
have temperature 1. ​ The potential followup at G9, from White getting all 6
of these moves, does not decrease the temperature for any of these 6 moves.
On this board, if the lower-right was immortal, then there would be only 3 moves
with temperature at least 1 that Black could play other than in the A,E,F components.
With at most 3 tenukies by Black, White can’t get more than 4 of ​ J5,J6,G5,D9,E9,F9 ,
before one of those 6 becomes what Black would play even if the upper-right was immortal.

For Chinese rules, the issue with the lower-right would
be too complicated for me to analyze in this post.
However, for Japanese rules - and more generally, for territory
scoring where pass-pass lifts basic ko-bans - I can show that
the non-immortality of the lower-right actually matters here:

Once there’s nothing else to be played, if Black has all of F4,F3,F2
and a ko threat, then White will owe a protective play in the lower-right:

Black plays J2. ​ If White doesn’t take and win the ko at this point - spending 2 moves
to Black’s 1 - then when Black has H1, Black plays J3 and uses Black’s ko threat.

The B,C,D components add up to 2 points for Black, and if Black plays first
in their sum, then Black can get that and have D7 be black rather than white.
If Black gets all of ​ F4,F3,F2,D7 , ​ then when White’s left group
has no outside liberties, Black will have a ko threat at A8.

Thus, White winning what would’ve been the chlled game if the lower-right were immortal,
at the cost of Black getting all of ​ F4,F3,F2 , ​ is not good-enough for White,
since Black will get D7 along with the expected 2 points from B+C+D, after which
White will owe a protective play in what otherwise would’ve been white territory.

Accordingly, one can treat ​ the result of blackF4 ​ as a miny. ​ ​ ​ Miny at Sensei's Library

When treated that way, the G component
Atomic Weight at Sensei's Library
gives only 1 atomic weight to White.

The E component is a *, so it has atomic weight 0.
The F component is a UP, so it gives 1 atomic weight to Black.
The B,C,D components have temperature strictly
less than 1, so they have 0 atomic weight.

The A component, and the result of White playing once in it, each favor Black.
Furthermore, the result of White playing twice in that component
is a *, so the A component gives 2 atomic weight to Black.

Thus the atomic weight favors Black by 2,
Atomic Weight at Sensei's Library
so Black wins no matter whose turn it is.

There are subtleties indeed that connect separate parts of the board. I think when you treat certain moves as sente, which you would do with the individual counting, then those go away, since it’s not going to be the case that white gets all of

and

Right as you mention

So you’d really have to add a lot of small endgame, like really a lot, in order to change the values slightly?

But even then does it change? Becuase is there not some sente hypothesis meaning that the options where black or white tenuki so many times are dominated or reversible or something?

It’ll take me a while to read through the whole post to see how you’re explaining why they could matter, but while they’re part of a bigger problem they are analysed separately the same way

but you are right that in the way (d) is combined into the whole board problem there is no ko and the black group is not under any threat

so maybe that’s an implicit assumption, I’ll look and see if the groups being immortal is mentioned somewhere.

So in that sense, what you mean is that if white moves first they can only get a tie in score? That’s being treated as a win for Black? E.g. something like

I guess the idea is that if white plays they can only turn the infinitesimal parts to 0, but black can turn it to ↑ which can be rounded to a full point is it?

“So you’d really have to add a lot of small endgame,
like really a lot, in order to change the values slightly?”

Yes

“But even then does it change?”

It wouldn’t for these, due to the behaviour of E+F I mentioned in part (b), but in general
it could reduce the atomic weight by 1, due to making Black reply 1 move sooner.
(In such a case, the difference would be approximately an UP.)

“Becuase is there not some sente hypothesis meaning that the options where black
black or white tenuki so many times are dominated or reversible or something?”

Yes, there is not such a hypothesis: ​ In a fight for the last move, defending a
corridor of length 29 is strictly better than defending a corridor of length 27.

This works in the same way as how, in a capturing race, having exactly 29
outside liberties is strictly better than having exactly 27 outside liberties.

“but while they’re part of a bigger problem they are analysed separately the same way”

because in the bigger problem, the White group was connected to more white stones
with a huge amount of eyespace. ​ ​ ​ ​ ​ ​ ​ For the reduction, one would simply note
“these stones are immortal” ​ , ​ ​ ​ since the reduced diagram won’t show that.

“So in that sense, what you mean is that if white moves first they
can only get a tie in score? That’s being treated as a win for Black?”

Yes; that’s what I meant.

“I guess the idea is that if white plays they can only turn the infinitesimal parts
to 0, but black can turn it to ↑ which can be rounded to a full point is it?”

No, since White responding in the UP reverses Black’s move to UP.
Due to that reversal, UP+* simplifies to
White moves to 0 ​ , ​ Black moves to 0 or *
, ​ so Black’s advantage in the sum consists of the fact that Black has a choice
and White doesn’t, not in anything else about what each player can move to.

The idea is actually that the weakness in White’s lower-right effectively
makes the F-column corridor almost 1 move shorter than it appears:
It only takes Black 1 net move - rather than 2 - to make that corridor no longer favor White.

(If you want to change your diagram to make it work in the intended way as a full-board puzzle,
then my suggestion would be replacing ​ blackH1,whiteH2 ​ with ​ blackJ3,whiteJ2 ,
since the latter is the pair which makes 2 eyes in the rectangular-six-in-the-corner
and leaves the most ko material while doing so, which I interpret as coming “closest”
to leaving an exploitable weakness without actually leaving such a weakness.)

That makes sense yes, I don’t think it’s mentioned, but I suppose you’re supposed to infer it from the groups in the combined diagram.

Just confirming I understand

If it wasn’t Go, and the position was ↑*, then white should change ↑ to *, and then the position is *+* =0. That’s what I mean.

This is ↑* in a game of Hadron (either player can play when there’s equal numbers of neighbours of the same colour, 0 of each, 1 of each, 2 of each - red can only take the top left to * and the upper right is *)

Like I understand that practically

Black having a choice is the advantage, and technically that ↑*={*,↑|0,↑} simplifies to {0,*|0} but it’s still a first player win, not a Black (Blue win).

so that seems like an oversimplification. Of course, all combinatorial games are in fact precisely about what each player can move to. Everything else is technical ways to simplify and evaluate quickly.

Right so white can’t just mirror the play exactly because the ko either can’t be won, or if it can white has to spend a move inside to finish it and can’t get the last move outside.

So if we want to adjust the puzzle to actually reflect the values that they would get if the groups were immortal then something like this you say:

and now there is a way for white to win by 1 point by playing first at say J5?

“Of course, all combinatorial games are in fact
precisely about what each player can move to.”

Yes, but that can easily be something else - that is, other than the players
having a different number of options - about what each player can move to.

The canonical form of chilling the A region is an example:
From that, each player has exactly 1 option, and yet Black is favored
because of ​ ​ ​ what Black’s move is to ​ ​ ​ vs ​ ​ ​ what White’s move is to ​ .

(I think this is useful because it at least suggests that the atomic weight is 0 or 1:
In fact, I’m not sure there are any examples from which
each move is to a symmetric game ​ ​ ​ but ​ ​ ​ the atomic weight is not in {-1,0,+1} ​ .)

“So if we want … at say J5?”

Yes, although I now realize that J5 would still not be an only winning move:
There would be one other move that also wins for White.

Mathematical Go: Chilling Gets the Last Point" by Elwyn Berlekamp and David Wolfe is a unique book that explores the game of Go through the lens of combinatorial game theory. It focuses on mathematical strategies and concepts, such as the “chilling” rule, to determine who gets the last point in various endgame positions. It’s a fascinating read for those interested in both Go and advanced mathematics.