Elsewhere here there has been a query on counting kos, and there has also been a little flurry of activity with book reviews. These two things prompted me to start this thread. It may well run into the sand, but if it doesn't it will probably prove valuable as a joint effort.

The background first. There are several ways to count boundary plays (yose). Except perhaps for those gifted with an affinity for numbers, the normal presentations range from awful to very, very awful. There are miai and deiri, terms for which a Japanese-English dictionary is little help (that's because they are technical accountancy terms). Miai has the extra problem of being associated with miai (nuff said!). When you use one and not the other, and why, is a mystery worthy of the tarot card aficionados. None of the presentations represent exactly what pros use - there are enigmas and variations, although deiri and miai will usually provide the basic beat. Mathematicians have their own take - CGT - which comes with an implicit elf'n'safety warning: don't try this at home. Some of us have ignored the warning and still bear the scars.

But every game of go has boundary plays, and so it is clearly a very, very important topic. So important that a pro will normally reckon to gain back about 40% of the value of handicap stones in this phase of the game. For us amateurs, a tiny improvement in your counting can save you five points a game instantly - half a stone! Kerchinggg! The only problem is, all this miai and deiri stuff gets in the way. Gedoingggg!

What to do? Well, O Meien has addressed the problem in a book called Boundary Plays - Absolute Counting. I thought we might look at this book in some detail, over a longish period, on the following basis.

1. I will present some ideas from the book, presented though in my own way. This will be in small dollops of honey for Pooh Bears like me of little mathematical brain, although when numbers are not around I will probably act more like Eeyore, perhaps a more suitable persona for someone with my L19 history.

2. Readers will attempt to understand what is written. Those who don't understand will query, and OTHER readers who have understood will answer. This is very important. I want to be Eeyorishly lazy, of course, but different ways of explaining things are rarely a bad thing.

3. There will be Tiggers among us - the bouncy, exuberant numbers people who understand it all instantly and upside down and who will be bursting to tell us why square pies are better than circles. I will have to ask them: please, please do come to the party but can you please remember Eeyore is listening, too. Above all, we want to limit this to O Meien's method.

4. If I judge we are staying on track and there is sufficient interest, I will move on to the next part. If we get to the end, what you will have obtained is a very useful, practical method for counting games and boundary plays that O himself claims to use. Even I understand parts of it, even if I sometimes have to take my socks off for the bigger numbers.

5. I will be excerpting the book. Those who wish to have many more examples and deeper explanations may wish to procure the original book and follow along that way. The book is:

ヨセ・絶対計算 by O Meien, Mainichi Communications (MyCom), 2004. About 1300 yen. ISBN 4-8399-1508-3.

Those interested in the subject enough to buy this book may also wish to order another one at the same time, to save on shipping. We may refer to it later on, but it is rather advanced and experimental, in that it is about miai and deiri counting but talks about them in a new theoretical way that shows how they relate to each other, but it also introduces new concepts to plug the gaps that these theories have. Despite that description, it is written in a clear and accessible way, with many examples to test yourself along the way. This book is:

目数小事典 (Compact Dictionary of Counting) by the Editorial Staff of the magazine Igo (always a great recommendation) and published by Seibundo Shinkosha, 2004. About 2000 yen. ISBN 4-416-70461-5. There is a companion volume to this but it adds nothing to the theory - it's just a workbook of problems.

The first excerpt has been put separately below for tidiness. I will go on to the next episode once I judge everyone has mastered the latest one (and wants to go on, of course - please let me know whenever it is time to stop).

If you follow O Meien's method, you will end up being to able to partition and count a game at the endgame stage as follows (Diagram 1), and devise a clear strategy for continuing.

Click Here To Show Diagram Code

[go]$$B Diagram 1 - Page 124
$$ ---------------------------------------
$$ | M M M M M M M M M M M . . X . T T T T |
$$ | M X O O M M M M M M M . O O X X O X T |
$$ | M X X O M M M O M M M M M O X T X T T |
$$ | O X O O M M X O . O M M X O O X T T T |
$$ | O O O M . O O X X O X M M . . O X T T |
$$ | O X O O . O X . X X O X M M O X X T T |
$$ | X X . . X X X X O O O M M M O . . T T |
$$ | T X O O . . O . X O X M M M . . . T T |
$$ | X X O . X T X . X O M M O O O X X X T |
$$ | T X O , . . X O O O X X O X X O O O . |
$$ | X O O O X X . . . M M O X X X X X X . |
$$ | T X X O O O X X . O . O O X O . X O . |
$$ | T X O O X X O X . . . . M O . . . O . |
$$ | T T X X O X . . O . . . O . O O . M M |
$$ | T T X O O X O X . . . X X . X O . O . |
$$ | T T X O X X X . O , O X T X X , X . . |
$$ | T T X O O X . O M M M O X T T X . . . |
$$ | T T X X O . X O . M O M O X T X T . . |
$$ | T T T T T T . . . . M M O O X T T T T |
$$ ---------------------------------------[/go]

In this diagram, Black has 83⅔ points (including 5 prisoners) and White has 80 (including 1 prisoner). Check first that you understand and agree with this count on the basis of the triangled stones as certain territory for Black plus dead white stones on the board, and Xs and dead black stones for White. Ignore the two-thirds bit, although for the curious it relates to the white stone below the top-right 4-4 point. Don't worry about the unmarked points, although, again for the curious, these are points excluded on the basis of being taken away by plays for one side AND plays for the other, one side going first and then the other. In other words, these are the still contestable points. Ignore also who is to move.

Click Here To Show Diagram Code

[go]$$ Diagram 2 - Page 19
$$ . . . . . . . . . . |
$$ , . . . . . , . . . |
$$ . . . . . . . . . . |
$$ . . . . . . . . . . |
$$ . . . . . . . . O . |
$$ . . . . . . . O . . |
$$ . . . . . . . . . O |
$$ , . . . . . , O O . |
$$ . . . X . X . X O O |
$$ . . X . X . . X X . |
$$ . . . . . . . . X O |
$$ --------------------[/go]

OM (O Meien) begins with a section on how to count unresolved areas. He means the sort of areas as in the extreme corner of Diagram 2. He has five sub-sections. The first sub-sections covers the case of simple atari or connection, where only one side can gain points, as here.

The examples relate to counting only Black's territory. The approach is as follows. In Diagram 2, either Black or White can get to play at 'a'. If Black gets there first, he makes 2 points of territory (i.e. counting 1 also for the prisoner) in the extreme corner. White gets nothing. If White plays 'a' first, White still gets nothing, but now Black also gets nothing.

In general terms, Black has 50% chance of getting 'a'. When totting up his overall territory, he can therefore conveniently count the corner area as 50% chance of getting 2 points, i.e. 1 point.

This may seem trivial, but according to OM, over 80% of people, including dan players, tell him that this area is unresolved and so is uncountable.

In fact, two possible things can be counted. Territory can be counted, and the value of a move can be counted. Here he is counting territory. The next chapter is on the value of a move, but the two things are apparently related.

Click Here To Show Diagram Code

[go]$$ Diagram 3 - Page 24
$$ . . . . . . . . . . |
$$ , . . . . . , . . . |
$$ . . . . . . . . . . |
$$ . . . . . . . . O . |
$$ . . . . . . . O . . |
$$ . . . . . . . . O . |
$$ . . . . . . . O . . |
$$ , . . . . . , . O O |
$$ . . . X . X . X X O |
$$ . . X . X . . X O a |
$$ . . . . . . . X O . |
$$ --------------------[/go]

OM's second section relates to the case where both sides can make some territory. It is only superficially different from the previous atari/connection type, but it is useful to see the steps worked out.

In Diagram 3, if Black plays 'a', he captures two stones and gets an empty point in the corner, a total of 5 points. Even if White captures the stone at 'a', Black takes three stones in a snapback at 'a', and he still gets 5 points (three captures minus the stone of his that was captured - and, for the record, note that we are always counting Japanese style here).

If White connects at 'a', he not only saves his stones but he gets 1 point in the corner.

So, when totting up Black's corner territory, assuming again equal chances for both players of getting to 'a' first, Black counts what? Not 50% of 5, but 50%, or half, of 5 minus the 1 point White got. That is (5 - 1) ÷ 2. Black's territory in the still unresolved corner in Diagram 3 is therefore counted as 2 points.

OM's next section extends this situation further to include dangly bits. Diagram 4 is an example. We will look at that in Part 2, but you may wish to try counting Black's territory

Click Here To Show Diagram Code

[go]$$ Diagram 4 - Page 27
$$ . . . . . . . . . . |
$$ , . . . . . , . . . |
$$ . . . . . . . O . . |
$$ . . . . . . . . . . |
$$ . . . . . . . O . . |
$$ . . . . . . . . . . |
$$ . . . . . . . O O O |
$$ , . . . . . , . X . |
$$ . . . . X . X . X O |
$$ . . . . . . . . X . |
$$ . . . . . . . . X O |
$$ --------------------[/go]

That concludes OM's involvement in this part. We now switch to Eeyore.

What Eeyore would like to know, is why, in Diagram 3, we are told to take the approach of counting Black territory as 2 points rather than taking the approach of saying Black has 2.5 points (half chances of getting 5 points) and White has 0.5 point (half chances of getting 1 point). I am assuming that the former case is absolute and the latter is relative. The book is about relative counting, of course, but - unless it was a far from impossible case of in one ear and out the other - it didn't register with me anywhere that OM explained why it has to be absolute. I am quite happy to take it on trust for the time being, but I would feel happier if I simply had confirmation that my suspicion about absolute and relative here is correct.

I plan to move fairly briskly from part to part unless someone asks for the brakes to be applied (and since this is intended for kyu players, lower grades should not hesitate to ask).

PS Can anybody please explain how to get rid of the junk in the caption below each diagram?

First excerpt and taking my socks off isn't even enough.

But with a lot of help from a calculator, I think I can say I know why these are 83 and 80.

_________________
Occupy Babel!

Excellent idea, John.



For non-ko situations (and for many ko situations), CGT, miai counting, and Absolute Counting (TM) give the same results.




I agree.



I would say that deiri gets in the way. It has its uses, but people tend to think about deiri values as if they were miai values. That does not work.



Excellent!

A little background of my own: I introduced the terms, miai counting and deiri counting on rec.games.go several years ago to counter the confusion caused by deiri counting, which was all most Western go players knew about. The trouble is that most of them thought that if you play a gote with a (deiri) value of 2 points, that you gain 2 points. You do not. You gain only 1 point, which is the miai value. Unfortunately, the new terms seemed to generate some new confusion, or perhaps they simply revealed previously hidden confusion.

After reading O Meien's book, I stopped referring to miai counting, and I thought that introducing a new term, Absolute Counting, might not be helpful, either, I simply started talking about how much moves gain. (To be precise, we should ask how much they gain on average.)

O Meien's book also represents a step forward in terms of another term that has caused much confusion, double sente. He does not evaluate double sente at all. In fact, he does not evaluate sente, but reverse sente. That is eminently correct, as it is reverse sente that gains something. (Remember the proverb, Sente gains nothing. Technically, it does gain something, but that something is not some number of points. Playing sente is like cashing a check. You get what is "rightfully" yours. )

Many thanks to John for excerpting this book for a Western audience.

_________________
The Adkins Principle:
At some point, doesn't thinking have to go on?
— Winona Adkins

Visualize whirled peas.

Everything with love. Stay safe.

An article series by John...commence drooling! I'll be reading this one with great interest.

I think we have to seriously consider putting a Sensei's Library page on "The Best of John Fairbairn", which puts links to his collected wisdom in one place.

With you as far as counting the points goes John (at least I think so). I don't understand the distribution of disputed territory in the lower right, but hopefully we'll come to that later.

It took a couple of re-countings, but I finally reached the correct tally (I forgot to count the stones already captured ).

I am however also unsure as to why S1&T1 are considered certain territory, but I suppose that will be cleared up sooner or later.

In any case, I'm looking forward to .... whatever comes. Thanks John!

_________________
Patience, grasshopper.

For those coming in at the end of the thread, I have added stuff to my second post. It was interrupted by She Who Must Be Obeyed.

Try to prevent using "(" and ")". Replace these for example with "-".

Click Here To Show Diagram Code

[go]$$ Diagram 4 -page 27-
$$ . . . . . . . . . . |
$$ , . . . . . , . . . |
$$ . . . . . . . O . . |
$$ . . . . . . . . . . |
$$ . . . . . . . O . . |
$$ . . . . . . . . . . |
$$ . . . . . . . O O O |
$$ , . . . . . , . X . |
$$ . . . . X . X . X O |
$$ . . . . . . . . X . |
$$ . . . . . . . . X O |
$$ --------------------[/go]

_________________
The really most difficult Go problem ever: https://igohatsuyoron120.de/index.htm
Igo Hatsuyōron #120 (really solved by KataGo)

That junk would be the brackets you used. I would suggest using komma (or something similar)

Click Here To Show Diagram Code

[go]$$c Something Something, page somewhere
$$ . . . . . . . . .
$$ . . . . . . . . .
$$ . X . X . X X X .
$$ . X X X . . X . .
$$ . X . X . X X X .
$$ . . . . . . . . .
$$ . . . . . . . . .[/go]

instead of

Click Here To Show Diagram Code

[go]$$c Something Something (page somewhere)
$$ . . . . . . . . .
$$ . . . . . . . . .
$$ . X . X . X X X .
$$ . X X X . . X . .
$$ . X . X . X X X .
$$ . . . . . . . . .
$$ . . . . . . . . .[/go]

(click on "Click Here To Show Diagram Code" if you are unsure what I mean)

_________________
While I was teaching the game to a friend of mine, my mother from the other room:
"Cutting? Killing? Poking out eyes? What the hell are you playing?"

Thanks, Cassandra/Mnemonic. It worked.

A perfectly logical position. You can reply that we can estimate the corner territory as 1 point for Black. That is true as far as it goes, and is practical.



OC, that is also a perfectly valid way of looking at it.

In other games we may say, Black won, 61 to 60, but in go we say, Black won by 1 point. <shrug>



I do not know why O Meien decided to call his method Absolute Counting (TM). But it sounds cool, and miai counting was already taken.

_________________
The Adkins Principle:
At some point, doesn't thinking have to go on?
— Winona Adkins

Visualize whirled peas.

Everything with love. Stay safe.

Click Here To Show Diagram Code

[go]$$ Diagram 2 - Page 19
$$ . . . . . . . . . . |
$$ , . . . . . , . . . |
$$ . . . . . . . . . . |
$$ . . . . . . . . . . |
$$ . . . . . . . . O . |
$$ . . . . . . . O . . |
$$ . . . . . . . . . O |
$$ , . . . . . , O O . |
$$ . . . X . X C X O O |
$$ . . X . X . . X X a |
$$ . . . . . . . . X O |
$$ --------------------[/go]

The circled point and 'a' seem to be related to each other (for instance, if white plays 'a', he can play at without getting captured). Oh well, it's beside the point, but it caught my eye.


I think your explanation is correct. Let's take the notation that black's points are positive and white's points are negative. 2.5 + (-0.5) = 2.0, which is what we should expect. The count in the area is two points (for black).

_________________
We don't know who we are; we don't know where we are.
Each of us woke up one moment and here we were in the darkness.
We're nameless things with no memory; no knowledge of what went before,
No understanding of what is now, no knowledge of what will be.

Thanks, John! This is really good stuff that I may want to read through again to make sure I fully understand.

My guess for diagram 4:

_________________
Someday I want to be strong enough to earn KGS[-].

After several recounts I arrived at 83 and 80, maybe by accident. They should invent a machine to do that for me. Wisely I stop recounting now
My reasoning was the same as CT's. Even found the same number, without cheating.

In the notation used so far, I suppose the calculation for diagram 4 would be:

This is very interesting, John! I especially like that you want to encourage people to discuss. I admit, though, I'm more of a Tigger than an Eeyore ...

I think I can follow the thought process.

For those having trouble counting the first diagram, I suggest trying Steve Fawthorp's counting lessons here. It's taken me a few times going through them, but it's started to work in my brain, and some of the meat of those lessons made it easier for me to succeed in the count.

Diagrams 2 and 3 sem to build a foundation for counting contested areas, which is interesting and new to me.

My own spin on describing this counting:

In both diagrams, the possibilities all revolve around a single spot on the board. Two possibilities: a) Black plays there, or; b) White plays there. OM seems to be saying that you can count that area as an average of the possible black points (using negative points for white).

So, for Diagram 2:

Black plays: 2 points
White plays: -0 points

Since there are only those two outcomes, we get (2-0)/2=1.

For Diagram 3:

Black plays: 5 points
White plays: -1 points

Average becomes (5-1)/2=4.

Diagram 4 is interesting. I'll copy it here:

Click Here To Show Diagram Code

[go]$$ Diagram 4 - Page 27
$$ . . . . . . . . . . |
$$ , . . . . . , . . . |
$$ . . . . . . . O . . |
$$ . . . . . . . . . . |
$$ . . . . . . . O . . |
$$ . . . . . . . . . . |
$$ . . . . . . . O O O |
$$ , . . . . . , . X a |
$$ . . . . X . X . X O |
$$ . . . . . . . . X b |
$$ . . . . . . . . X O |
$$ --------------------[/go]

This situation revolves around two points, not one. My guess is that we'll be able to apply the same idea here. Let me try it. EDIT: Hidden, just like mitsun did, in case people want to try before reading.

This is not a practical way to find the value of the corner region, but a demonstration that the value people have found is correct.



OC, the demonstration depends upon the play being correct.

_________________
The Adkins Principle:
At some point, doesn't thinking have to go on?
— Winona Adkins

Visualize whirled peas.

Everything with love. Stay safe.

While it may not be generally practical, in this fortunate example where there are exactly 4 possible outcomes, your diagram completely and excellently unfuddles the issue.

_________________
Patience, grasshopper.

Does it really matter? 2.5:0.5 and 2:0 is the same, in both cases black wins by 2 points. And since it doesn't really matter, it makes sense to prefer the version where you only have to remember one number instead of two