After, for some of us, a tough session in Part 2, this is an easy bit - just tidying up a loose end and providing a summary. For those following along with the book, we are here dealing with pages 32 to 39.

The calculations we have seen so far ended in a result that gave a value for counting a territory that ended in a whole number. The value could just as easily end in a fraction, but the method and the thinking behind it are exactly the same.

What I take from the discussion so far, is that there is no really good term for what this value represents, but Expected Value struck me as best of the bunch, and I propose to use it but with cap E V to show it is being used in a slightly weird (to me) sense.

Expected Value for a player represents the size of a territory based on the assumption that he has a 50-50 chance of actually getting it. The actual calculation to get that value seems to be fiddly even for numbers experts and there is more than one way to do the calculation (and the various methods apparently have no name), but I liked Robert Jasiek's explanation and especially his derivation of the formula (B + W) ÷ 2, which is the one OM uses (but does not explain, as RJ did). From now on, rather than using my own ponderous approach I will just use that formula.

For those, who like me, feel a bit like those climbing a rock-face only thanks to a reassuring but paradoxically still very worrying safety rope, I would say this: this is perhaps as tough as it gets. The calculation of the figures is fiddly even for the experts. But it is also mechanical and therefore easy enough to do. Furthermore, even if you are still (like me) unsure where the figures really come from, it is perfectly OK to accept them on trust, as they are still useful - as I hope this section will show.

Click Here To Show Diagram Code

[go]$$ Diagram 6 - Page 32
$$ . . . . . . . . . . |
$$ , . . . . . , . . . |
$$ . . . . . . . . . . |
$$ . . . . . . . . . . |
$$ . . . . . . . . O . |
$$ . . . . . . . O . . |
$$ . . . . . . . . . O |
$$ , . . . . . , O O . |
$$ . . . X . X . X O O |
$$ . . X . X . . X X a |
$$ . . . . . . . . X . |
$$ --------------------[/go]

In Diagram 6, Black can play 'a' and get 1 secure point. White can play 'a' and take that point away. Applying the 50-50 formula to this as before we get (B + W) ÷ 2 = (1 + 0) ÷ 2 and so the Expected Value for Black here is 0.5. It's only the fact that result is a fraction that is worthy of comment.

But for completeness, OM also gives examples with captures and dangly bits. Here are a couple.

Click Here To Show Diagram Code

[go]$$ Diagram 7 - Page 33
$$ . . . . . . . . . . |
$$ , . . . . . , . . . |
$$ . . . . . . . . . . |
$$ . . . . . . . . . . |
$$ . . . . . . . . . . |
$$ . . . . . . . . . . |
$$ . . . . . . . O O O |
$$ , . . . . . , . X a |
$$ . . . . . X . . X O |
$$ . . . . . . . . X O |
$$ . . . . . . . . X . |
$$ --------------------[/go]

Here the Expected Value for Black is (5 + 0) ÷ 2 = 2.5.

Click Here To Show Diagram Code

[go]$$ Diagram 8 - Page 35
$$ . . . . . . . . . . |
$$ , . . . . . , . . . |
$$ . . . . . . . . . . |
$$ . . . . . . . . . . |
$$ . . . . . . . . . . |
$$ . . . . . . O O O O |
$$ . . . . . . . X X a |
$$ , . . . . . , X O O |
$$ . . X . X . . X X O |
$$ . . . . . . . X O X |
$$ . . . . . . . X b c |
$$ --------------------[/go]

This is one of those cases where iteration or recursion is required. The B in (B + W) ÷ 2 is 10 (i.e. Black plays at 'a' and gets 10 points of secure territory - play in the position is now finished). For the W, i.e. when White gets to play at 'a', we need to sort out first the result of the dangly bit, i.e. make a subsidiary calculation of the 50-50 chances of who gets to play 'b' or 'c' first. On that basis the formula boils down to (10 + 1) ÷ 2 = 5.5. Again, it's only the fraction that is being highlighted.

At this stage, OM gives a summary in a box. He says:

"The method of counting territory which has an unresolved portion is derived from obtaining the numbers of points for the respective cases when both Black and White have played there and, if the possibility of these values is 50-50, halving them."

He then adds a section to make what he says is a "truly fundamental, important point", namely that in Diagram 9 "The Black territory at the 1-1 point in the corner is 1 point, and the important thing is not only that this is regarded as 1 point but that it is acknowledged as being exactly the same as 1 pure point as in Diagram 10."

Click Here To Show Diagram Code

[go]$$ Diagram 9 - Page 37
$$ . . . . . . . . . . |
$$ , . . . . . , . . . |
$$ . . . . . . . . . . |
$$ . . . . . . . . . . |
$$ . . . . . . . . . . |
$$ . . . . . . . . . . |
$$ . . . . . . . . . . |
$$ , . . . . . , O O . |
$$ . . . . . X . X O O |
$$ . . . . . . . X X . |
$$ . . . . . . . . X O |
$$ --------------------[/go]

Click Here To Show Diagram Code

[go]$$ Diagram 10 - Page 37
$$ . . . . . . . . . . |
$$ , . . . . . , . . . |
$$ . . . . . . . . . . |
$$ . . . . . . . . . . |
$$ . . . . . . . . . . |
$$ . . . . . . . . . . |
$$ . . . . . . . . . . |
$$ , . . . . . , O O . |
$$ . . . . . X . X O O |
$$ . . . . . . . X X X |
$$ . . . . . . . . X . |
$$ --------------------[/go]

He goes on to stress this in a few more different ways, pointing out that it is not a "convenience" figure, that it is "an aid to evaluating moves", and that "amateurs don't understand this very well". And since we clearly are assumed not to have got the point even then, he gives a catalogue of positions where one is "exactly the same" as the other. I will give two of the examples, one easy, one tricky. In each case (Diagrams 11 and 12), Black's territory on one side of the board is the same as on the other.

Click Here To Show Diagram Code

[go]$$ Diagram 11 - Page 38
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . , . . . . . , . . . . . , . . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . O . . |
$$ | O O O O . . . . . . . . . . . . . O O |
$$ | X X X O . . . . . , . . . . X , X X O |
$$ | . . X O . . . . . . . . . . . . X O . |
$$ | X . X O . . . . . . . . . . . . X O O |
$$ | . . X O . . . . . . . . . . . . X O O |
$$ ----------------------------------------[/go]

Click Here To Show Diagram Code

[go]$$ Diagram 12 - Page 39
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . , . . . . . , . . . . . , . . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | . . . . . . . . . . . . . . . . . . . |
$$ | O O O O O . . . . . . . . . . O O O O |
$$ | O X X X O . . . . . . . . . . . X X . |
$$ | X X . , O . . . . , . . . . . , X O O |
$$ | . . X X O . . . . . . . . . . . X X O |
$$ | X . X O O . . . . . . . . . . . X O X |
$$ | . . X O . . . . . . . . . . . . X . . |
$$ ----------------------------------------[/go]

In Diagram 11, both positions count as 5 points. In Diagram 12, both positions count as 5.5 points.

The next section, though still part of Chapter 1, is on counting the resolved area inside groups and determining the fuzzy areas between opposing groups. Although it makes some reference to Expected Values, it is mostly a number-free zone and so is a distinct change of pace.

I suggest, therefore, that if there is anyone still with questions on Expected Values, they should ask our experts now, before we move on.

Indeed. In what sense are the two diagrams "exactly the same?" In the sense that when counting territory to see who is ahead, it makes no difference if the bottom right corner looks like diagram 9 or diagram 10? Why is this? Is it because there is a 50% chance in diagram 9 that black will make 2 points, and we assume that there are other similar situations on the board that will get divvied up?

_________________
Patience, grasshopper.

I don't know if this was the case or not for you, but I think the point is, it's important to not see diagram 10 as having captured a stone. It's comparing 1 point of territory with "either 2 or 0". These threads are making for some interesting reading

I have the same sort of questions and come to the same sort of conclusion. It is not explained by OM - or at least not yet. This back-to-front way of writing is pretty typical of Japanese authors, but you get used to it. However, when it's compounded with numbers as well, it really makes my brain hurt.

I think it's only *exactly* the same when there are enough moves of similar size around:

Click Here To Show Diagram Code

[go]$$B Standard 9x9 board
$$ -------------------
$$ | . . . . . . . . . |
$$ | . . . . . . . . . |
$$ | . . , . . . , . . |
$$ | . . . . . . . . . |
$$ | . . . . , . . . . |
$$ | X X X X O O O O O |
$$ | . X , a . O , X O |
$$ | X X X O O O . X b |
$$ | . . X X X X X X O |
$$ -------------------[/go]

On this board black has *exactly* 4 points on the bottom. There's no way for it to be anything else.

a has value (1 +(-1))/2 = 0
b has value (2 + 0)/2 = 1
plus the 3 resolved points.

Now a and b are miai, and if white takes one, black takes the other:
if white takes a, black takes b, leaving 3 + 2 - 1 = 4
If white takes b, black takes a, leaving 3 + 1 = 4

I guess there must be more to it and even on a much more general level but I forgot the details. Maybe it was already in "Winning Ways" and "On Numbers and Games"?

Yes.

I think it's also important to point out in what sense they are *not* exactly the same. In one, it is still possible to gain points. In the other, it's settled.



In a way, this is correct. If you had two copies of diagram 9, they would be miai, and black will make exactly 2 points in one copy, and exactly 0 in the other. So black made exactly 1 point per copy. That's the same as if there were two copies of diagram 10. There is no estimate here (of course that's always assuming correct play by both).

But it is also correct when there are no other same-sized moves available. If you have only a single copy of the position, and everything else is already settled, then black too has exactly 1 point. And there is still a move left that gains 1 point. Whoever makes that move will change the score by 1 point. But that hasn't happened yet.

This method always determines what the score is right now. It doesn't necessarily tell you who's winning, for that you have to consider the remaining moves, too. A trivial example: Let's say the count on the board is B+5. But white plays a tesuji, killing a 20 stone black group. Black resigns. Was the count of B+5 wrong before? No, the count was correct, but there was still a 20-point move left for white, with white to move. Diagram 9 is exactly the same, just that it's only 1 point instead of 20.

Of course, the 20-point example is a bit extreme, and in that case the count was not really useful. But usually, the remaining moves are of similar size, and the size decreases slowly, that is, when there are 10-point moves, there surely are a few 9-point moves, a few 8-point moves and so on. So in general, the count "right now" is a good approximation of the final count of the game. Note that this is the only point where the word "approximation" comes into play: The "current count" (i.e. what we have been talking about in these threads) is an approximation of the "final count" at the end of the game, when all remaining moves have been (correctly) played

Right - if you consistently get fewer than half of the moves worth a specific number of points, your opponent is dominating you. This is why tedomari is worth double, if I understand correctly.

There's a great java program called cgsuite that does these kinds of calculations automatically:
http://cgsuite.sourceforge.net/

To figure out the example in diagram 5, here's an example session:


It uses "Mean" where we've been saying "count". The syntax is


And it can build up more complicated positions from simpler ones, and does other fun CGT stuff like thermographs. It can add positions too, like my example above:

Unfortunately, I have mislaid my copy of OM's book. But let's admit a couple of things. First, as a 9 dan, he is perfectly aware of the difference between these two diagrams. Second, he may not be a mathematician or logician. Third, he is not aiming his book at mathematicians or logicians.

I suspect that he is not being completely fair to amateurs, some of whom may have logical or mathematical questions about how the 1 point in each diagram is the same. (The subtitle of the book is, as I recall, "Do you know what 1 point is?")

What we can certainly say is that the count in the corner of each diagram is exactly the same, 1 point. Exactly what that means involves logic and math. (I will be glad to speak to that later. ) The count has the nice property that if you add the gains and losses of each subsequent play to the overall count, in the end the result is equal to the net score. In the case of the first diagram, eventually we will add or subtract 1 point from the current count, depending on who plays in the corner. In the case of the second diagram, unless Black stupidly fills his own territory, we will not. The addition or subtraction will affect the eventual score, of course, but it will not affect the original count. That is -- or should be, IMO --, OM's point.

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

Visualize whirled peas.

Everything with love. Stay safe.

I still claim that this Absolute Value (as OM seems to call it) is really "the correct count", and is independant of other moves on the board. Or, as I should more accurately say, it is the count that makes the most sense, because unfinished games don't really have a "score" according to the rules of go.

I think most posters here agree that the score is correct when there are same-sized moves on the board, or when there are multiple copies of a position. Let me try to offer another perspective that tries to explain why that is always so, even without multiple copies:

Let's consider a simple example:

Click Here To Show Diagram Code

[go]$$ Diagram 7 - Page 33
$$ . . . . . . . . . . |
$$ , . . . . . , . . . |
$$ . . . . . . . . . . |
$$ . . . . . . . . . . |
$$ . . . . . . . . . . |
$$ . . . . . . . . . . |
$$ . . . . . . . O O O |
$$ , . . . . . , . X a |
$$ . . . . . X . . X O |
$$ . . . . . . . . X O |
$$ . . . . . . . . X . |
$$ --------------------[/go]

How many points does black have here?

It's clear that when white plays at 'a', black has 0 points. When black plays at 'a', black has 5 points. That's easy, but that was not the question. How many points does black have *now*?

Of course the answer to that is in reality entirely subjective, the rules of go don't define the score of unfinished games. We are free to choose any counting method we like, so let's explore several possible answers:

a) 0
Black doesn't have any points here yet.
That implies that a black play at 'a' gains 5 points. Ok so far.
But that would also imply that a white play at 'a' gains nothing, because after white plays at 'a', the count is still 0. But clearly playing at 'a' is worth something to white.

b) Can't be counted exactly since it isn't played out yet.
That is clearly unsatisfactory. Reading the whole game out to a position that can be scored according to the rules of go is not practical.

c) 2.5 (OMs answer)
Then the black move at 'a' is worth 2.5 (moves from a position worth 2.5 to a position worth 5).
And the white move at 'a' is also worth 2.5 (moves from a position worth 2.5 to a position worth 0).
That is the only answer where the black move and the white move are worth the same. That is a very desirable property, because it agrees with intuition. If there is a 10€ bill on the table and I take it, I gained 10€. If you take it, you gained 10€, too. There is no way that any amount of money can be on the table that's worth more when you take it compared to when I take it.


You can also play the same game with sequences of more than one move, you'll find that OM's count will always be the only count that satisfies the property that a move is always worth the same to both players. Of course it's somehow arbitrary to choose this as the defining property of our count. But the choice makes sense (at least to me ).

PS: Of course that's only looking at gote moves for now. I expect sente moves will be treated in a later chapter

One might define the current position's value (on all intersections or on a locale, i.e. here this means those local intersections currently considered for boundary play) to be successive passes followed by the score according to the rules.

Indeed.



This was my view before studying CGT. But the count is not arbitrary. If two of some position has a definite value of 3 points, it makes no sense to say that one of them is worth 2 points but the other one is worth only 1 point. Besides that, if the overall count is in your favor and you have the move, and there are no kos or future kos, then with correct play you will win. You cannot say that about mere estimates or expected values.

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

Visualize whirled peas.

Everything with love. Stay safe.

This is carrying the OM method too far. The method gives the expected value of a position, under the assumption that either side has an equal probability of playing there first, which in turn more-or-less assumes there are lots of similar sized moves left to play. If there is only one move left (or if one move is much larger than any other remaining moves), then there is no uncertainty about who will get to play there first, and the OM method is not applicable. The expected value of a position is most definitely not the same as the value of the next move in that position. But I suspect that is the topic for an upcoming chapter.
Actually, in this example, I would say the OM count is wrong, in the sense that it does not accurately evaluate the game position. The OM count should at least include the availability of the W tesuji, weighted by 50% if playing there is double gote. However if there is one move left worth much more than anything else, the OM count will not give the correct evaluation of the position. A better count in this particular situation would be obtained by weighting the W tesuji at 100%, since it is definitely going to be the next move. The standard OM count then might or might not be applicable to the rest of the game.

Bill, so "count" is the better term than "expected value"?

expected value deals with random var.
count is not.
so you are wrong.

_________________
"The more we think we know about
The greater the unknown"
Words by neil peart, music by geddy lee and alex lifeson

Count is more precise.

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

Visualize whirled peas.

Everything with love. Stay safe.

IMO expected value can be used on local situations because you need to average over all possible other board situations, which makes it basically a random variable. But I wouldn't talk about expectancy values in whole board situations.

Of course a deterministic result is the degenerate special case of a random variable

_________________
Sanity is for the weak.

Since we do not have random variables, Bill and I are right:)

Li Kao, looking at B or W moving first locally with 50% probability is a maybe convenient but also naive model. In reality we have deterministic game trees.