It seems better to start a new thread for each part so that latecomers can still contribute to earlier threads.

It is still early days on this exciting journey, but we have seen some sights already. We saw that it is possible to partition up a typical endgame position into resolved areas and unresolved areas. The resolved areas can be counted up fairly easily with enough fingers and toes. The unresolved areas are trickier. There are two things to consider there. One is how to mark out the edges of the area. Although they look fuzzy, they are not and a pretty simple method is used for the marking out.

The other issue is how to count these unresolved areas.

With the resolved areas we just made a count for each side: Black 83 and a bit, White 80. We can compare Black to White and say he has more - relative counting. We could subtract one score from the other and get a margin of victory (83-80 = 3) - absolute counting. For these areas it doesn't seem to matter which method of counting is used, but for the unresolved areas O Meien (OM) has chosen absolute counting, as per the title of his book. With luck, his reasons will become apparent later.

His method starts with learning to count the unresolved areas, and we shall be continuing with that in a moment, but two things have emerged so far that need, I think, some emphasis.

One is that miai and averages have been mentioned by other readers. This extra explanation is good, useful and intended, but these words are NOT part of OM's method, even if the ideas are implicit. Do keep that in mind.

The other is a bit of fuzziness about terminology of a different kind. When counting the unresolved areas, OM used the phrase "Black has a territory of 2 points". Not quite right, but that's a Japanese language issue we won't go into. Bill used a phrase saying that we "estimate Black's territory as 2 points". Not quite right either, for some of us, I suggest. It's quite possibly the correct usage in mathematics, but in ordinary speech "estimate" suggests vagueness (cf. the well known issue of estimate over quotation when dealing with tradesmen). OM is emphatic that his numbers are not an estimate, in that sense, but a precise calculation. However, we do need a word to remind us we are not dealing with actual or resolved territory, and this word seems as good as any so long as we remember the caveat.

We resume again with repetition of Diagram 4. Recall that so far (our Part 1) OM has treated simple unresolved corner positions where, depending on who moves first, (1) Black could make some secure territory (B) or none and White likewise makes some (W) or none, and that we count, or rather "estimate", this from Black's point of view, by OM's absolute counting method, as (B + W) ÷ 2. W here will always be 0 or a minus figure because we are looking at everything from Black's point of view. The division by two is to represent the fact that both players have a 50-50 chance of making the play.

Some of us are actually taking the applicability of this method on faith. It is also used in Diagram 4, which relates to the type of position with dangly bits.

Click Here To Show Diagram Code

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

Black can play at 'a' and get a secure territory in the extreme corner of 5 points, and White will have 0. White can also play at 'a'. He still gets 0, but he doesn't deprive Black of 5 points in this case, because a play at 'b' remains for both players. Here is how OM handles this: simply take Black's gains in two parts and add them together, but make some allowance for the fact that the play at 'b' is less likely to happen. With 'a' he makes 5 points but this figure is actual territory and has not yet been processed into an absolute value. But if White gets to play at 'a' instead, we have another position, where 'b' is the key move, and we count that differently: we take the "processed" value (obtained as in Part 1) of 1 point by the absolute counting method. Black's territory is therefore, for the purposes of the method is counted as 5 + 1 and not 5 + 2. White's territory can be counted in the same way but here it becomes 0 + 0.

For people like me, it is easy enough to understand that the second figure (1, relating to 'b') has to be treated differently from the 'a' figure, to cover the fact that it is lower down the food chain and less likely to be used. But the reason for treating it in this particular way is far from obvious - I'm still making a leap of faith. OM admits it is fiddly but claims you get used to it.

Anyway, once we get this 5+1 for B, the operation proceeds as before, i.e. (B + W) ÷ 2, which is here ((5+1) - (0+0)) ÷ 2, so that the position of Diagram 4 is "estimated" as being worth 3 points for Black by absolute counting.

This method would apply also to the case where there is a third or a fourth dangly bit, but mercifully OM treats these as dingleberries and ignores them, at least for the time being. But the case of an 'a' and a 'b' move where White also makes some territory merits a look (the same distinction as between the two examples in Part 1), and one OM example is as follows.

Click Here To Show Diagram Code

[go]$$ Diagram 5 - Page 30
$$ . . . . . . . . . . |
$$ , . . . . . , . . . |
$$ . . . . . . . O . . |
$$ . . . . . . . . . . |
$$ . . . . . . . . O . |
$$ . . . . . . . O . . |
$$ . . . . . . X X O O |
$$ , . . . . . , . X a |
$$ . . . . . . . X X O |
$$ . . . . . . . X O O |
$$ . . . . . . . X b X |
$$ --------------------[/go]

If Black gets to play 'a' in Diagram 5, he makes a secure territory of 7 points. If White gets to 'a' first, he gets no territory yet, but a move at 'b' remains. If we process that position relating to 'b' first, by absolute counting we estimate it from Black's point of view as worth -1, that is Black gets 0 but White gets 1 (i.e. -1 from Black's point of view).

Using that figure, we then process the 'a' position again as B + W ÷ 2, or (7+0) + (0+(-1)) ÷ 2. That is, Diagram 4 is also estimated as worth 3 to Black.

I will break off here as this portion does not seem easy to me, and a better explanation can almost certainly be offered by someone. As a matter of fact, the equations with lots of brackets above are my own attempt to understand what is going on. OM uses equations but in the last example he simpy has {7 +(-1)} ÷ 2. I have expanded it for my own benefit. I hope I haven't botched it.

Fortunately, this is as tricky as the counting bit gets. If we can get over this hump, we can coast for a while. The next section covers the case where the estimate ends in a half point, but that's a trivial step forward. There is then an easy set of examples where OM stresses that positions counted by his method and positions with already secure territory are equivalent. The second part of Chapter 1 of his book is devoted to how to count territory in resolved areas - not too hard actually, though you need to keep on your toes. That will take us to a third of the way through the book. Chapter 2 next is on "the value of a move". It applies Chapter 1 but puts sente and gote into the mix. Fun but tricky, and we'll need Bill on hand.

Chapter 3 is an extension of Chapter 2, adding kos and trades. Chapters 2 and 3 are maybe the toughest but are full of interest. This is real go.

Chapter 4 introduces an election night atmosphere. This is where you learn to declare who's first past the post before the vote count is even complete. Chapter 5 is very similar, but is a set of examples from actual games showing how all the elements are put together at strategic level.

I would use the term "Expected Value" (EV): The EV of black territory is 2 points.

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Click Here To Show Diagram Code

[go]$$ Diagram 5 - Page 30
$$ . . . . . . . . . . |
$$ , . . . . . , . . . |
$$ . . . . . . . O . . |
$$ . . . . . . . . . . |
$$ . . . . . . . . O . |
$$ . . . . . . . O . . |
$$ . . . . . . X X O O |
$$ , . . . . . , . X a |
$$ . . . . . . . . X O |
$$ . . . . . . . X O O |
$$ . . . . . . . X b X |
$$ --------------------[/go]

Hmmm, I had success with the last Diagram. Let's try it with this one.

We start with a. I find this is another interesting portion of the exercise. A bit of understanding of the position is needed to know which point is most important. If black decided (for whatever reason) to start with point b, the position is worth less (as Black now needs two moves to resolve the position instead of one).

Here's my calculation, hidden so others can try without reading it first.

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(emphasis by me)

I think that the bold part is important.

Let's try Diagram 4 by (what I understand to be) OMs method step by step:

1)

Click Here To Show Diagram Code

[go]$$
$$ . . . . . . . . . . |
$$ , . . . . . , . . . |
$$ . . . . . . . O . . |
$$ . . . . . . . . . . |
$$ . . . . . . . O . . |
$$ . . . . . . . . . . |
$$ . . . . . . . O O O |
$$ , . . . . . , . X 1 |
$$ . . . . X . X . X W |
$$ . . . . . . . . X C |
$$ . . . . . . . . X W |
$$ --------------------[/go]

Black has 5 points. White has 0 points. Easy.

2)

Click Here To Show Diagram Code

[go]$$W
$$ . . . . . . . . . . |
$$ , . . . . . , . . . |
$$ . . . . . . . O . . |
$$ . . . . . . . . . . |
$$ . . . . . . . O . . |
$$ . . . . . . . . . . |
$$ . . . . . . . O O O |
$$ , . . . . . , . X 1 |
$$ . . . . X . X . X O |
$$ . . . . . . . . X b |
$$ . . . . . . . . X O |
$$ --------------------[/go]

How many points do black and white have? It's still unresolved, so we need to apply the same method to this figure:

2a)

Click Here To Show Diagram Code

[go]$$
$$ . . . . . . . . . . |
$$ , . . . . . , . . . |
$$ . . . . . . . O . . |
$$ . . . . . . . . . . |
$$ . . . . . . . O . . |
$$ . . . . . . . . . . |
$$ . . . . . . . O O O |
$$ , . . . . . , . X O |
$$ . . . . X . X . X O |
$$ . . . . . . . . X 1 |
$$ . . . . . . . . X W |
$$ --------------------[/go]

Black has 2 points (1 territory, 1 prisoner), white has 0 points.

2b)

Click Here To Show Diagram Code

[go]$$W
$$ . . . . . . . . . . |
$$ , . . . . . , . . . |
$$ . . . . . . . O . . |
$$ . . . . . . . . . . |
$$ . . . . . . . O . . |
$$ . . . . . . . . . . |
$$ . . . . . . . O O O |
$$ , . . . . . , . X O |
$$ . . . . X . X . X O |
$$ . . . . . . . . X 1 |
$$ . . . . . . . . X O |
$$ --------------------[/go]

Black has 0 points, white has 0 points.

2c)
We sum that up: Black has 2 points in 2a) and 0 points in 2b). White has 0 points in both.
We use OMs formula: (B+W)/2 = ((2+0)-(0+0))/2 = 1.

So now we know:
2)

Click Here To Show Diagram Code

[go]$$W
$$ . . . . . . . . . . |
$$ , . . . . . , . . . |
$$ . . . . . . . O . . |
$$ . . . . . . . . . . |
$$ . . . . . . . O . . |
$$ . . . . . . . . . . |
$$ . . . . . . . O O O |
$$ , . . . . . , . X 1 |
$$ . . . . X . X . X O |
$$ . . . . . . . . X b |
$$ . . . . . . . . X O |
$$ --------------------[/go]

Black has 1 point, white has 0 points.

3)
Now the important part: The count by his method is equivalent to already secured territory. Black actually has 1 point here, and white has 0 points. We don't treat the move at b different, a more accurate description would be "we don't treat the move at all". We have already established a count for this position, and now we just treat it as if it were already secure territory, with no move left.

4)
Now we sum up again: Black has 5 points in 1), and 1 point in 2). White has 0 in both.
Again using the formula: ((5+1)-(0+0))/2 = 3.


What we effectively did at step 3) is that we switched diagram 2) with a different one, that has the same count, but where the territory is resolved.

Click Here To Show Diagram Code

[go]$$W
$$ . . . . . . . . . . |
$$ , . . . . . , . . . |
$$ . . . . . . . O . . |
$$ . . . . . . . . . . |
$$ . . . . . . . O . . |
$$ . . . . . . . . . . |
$$ . . . . . . . O O O |
$$ , . . . . . , . X O |
$$ . . . . X . X . X O |
$$ . . . . . . . . X X |
$$ . . . . . . . . X C |
$$ --------------------[/go]

Black has 1 point, white has 0 points. Note that this is very different from making the move at 'b' in the previous diagram. This is another, unrelated diagram, without a white prisoner in the corner.

Since we know that values arrived at using Absolute Counting are equivalent to settled positions (that at least we have to believe because OM said so ), we might as well just substitute the "fuzzy" situation with a perfectly resolved situation, and count that.

OM is quite right that these numbers are not mere estimates. As my demonstration in the previous thread shows, if you add four such positions together, their combined score is a constant. That does not happen with mere estimates. Professor Berlekamp has coined the term, count, for these values.

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— Winona Adkins

Visualize whirled peas.

Everything with love. Stay safe.

Hi, Eagle Eye!

I think that there is a typo in the diagram.

Click Here To Show Diagram Code

[go]$$ Diagram 5 - Page 30
$$ . . . . . . . . . . |
$$ , . . . . . , . . . |
$$ . . . . . . . O . . |
$$ . . . . . . . . . . |
$$ . . . . . . . . O . |
$$ . . . . . . . O . . |
$$ . . . . . . X X O O |
$$ , . . . . . , . X a |
$$ . . . . . . B . X O |
$$ . . . . . . . X O O |
$$ . . . . . . . X b X |
$$ --------------------[/go]

There. All better now.

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

Visualize whirled peas.

Everything with love. Stay safe.

MUCH better

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Even this is a typo.

Ô Meien has put the circled stone one point to the right.

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Igo Hatsuyōron #120 (really solved by KataGo)

I second this proposal; it makes sense both mathematically and conversationally. Another way of saying it, maybe more in line with the OM method: "this position has an EV of +2 points for Black".

Using flOvermind's diagrams and their numbers, I (and I think also many other players) use (besides other, alternative methods) the following method, for which I do not know the correct name and which for the moment I call "calculation of one local endgame count by iterative averages":

- The initial position depends on Black moving first (position (1)) versus White moving first (position (2)).
- In position (1), the count is 5 in favour of Black.
- Position (2) depends on Black moving first (position (2a)) versus White moving first (position (2b)).
- In position (2a), the count is 2 in favour of Black.
- In position (2b), the count is 0.
- The average of position (2) takes B from (2a) and W from (2b) and is (B-W)/2 = (2-0)/2 = 1.
- The average of the initial position takes B from (1) and W from (2), where the value 1 in favour of Black is negated as -1 to be expressed in favour of White, and is (B-W)/2 = (5 - (-1)) / 2 = 6 / 2 = 3.

(Alternatively and by convention, one might instead write (B+W)/2 and skip the double negation for White.)

This yields the same count as O Meien's method. The iterative averages method "miraculously" negates values when considered in favour of the opponent while O Meien's absolute counting method "miraculously" negates values when shifting them from White to Black's bracket of summonds or vice versa. I guess that there is an implied theorem of the two methods being equivalent. Does anyone have a general proof?

Or are the two methods just the same but a bit of further text study needs to reveal this yet?

One might even say that Black expects, or can expect to make 3 points of territory. With that, we're back to ordinary expressions.

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Howsabout: Black can count 3 points of territory.

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This is how I do it, and I think it's the way that makes the most sense. I admit that OM's explanation (or JF's explanation of OM's explanation) just confuses me, so I can't say whether the two methods are the same (but I hope they are!). I think "iterative averages" is a fine description, although I think of it as "recursive averages" personally.

Would it be better to say that our estimates are fuzzy, rather than vague? Vagueness implies something like the sorites problem (if I estimate +1 in this area it could just as easily be 0, and if I could estimate 0 it could just as well be -1...), whereas fuzziness implies that my expectations bear a somewhat loose relation to reality.

The key difference is that when I'm making an estimate, even if the estimate is fuzzy, I should expect my estimates to always be right on average. If not, it's a bad estimate. If my contractor's bill comes in way above his estimate every time, he's either an idiot or a liar. Similarly, as Bill says, if my estimates of the score in all the remaining contested areas are all correct (that is, if they are good estimates), then even if I estimated too high for Black here and too high for White there, I estimated the total score perfectly.



Chaining together the probabilities in this way is very standard. What are the odds of flipping two coins and having them both come up heads? What are the odds of just missing your train at Harvard Square, and then again at Park Street? What are the odds of rolling ten or higher on two dice?

To figure out the probability of any event that's composed of two independent events, you just multiply the probability of the first event by the probability of the second. (So you get 25%, 100%, 11%.) In this case, OM is suggesting that the probability of Black getting a move that's worth a specific number of points is 50%.

At my level, that's clearly a fiddly assumption But multiplying the probability of the follow-up by the probability of the initial move is the only thing to do.

Yes, the values of positions are technically fuzzy. You can also give them a probabilistic semantics.



I do not recognize what I said in that.



There is the practical value of the probabilistic semantics.

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

Visualize whirled peas.

Everything with love. Stay safe.

I think both methods for calculation presented here are equivalent. I would call them 'recursive':

The value V of a position is always the average of black's gain minus white's: V = (B - W)/2.
Let's assume point "a" gets played first and call the black/white gains Ba and Wa.

Black playing "a" gains 7 points immediately, so B(a) is 7.
White playing "a" leads to a board position which has to be further evaluated, but there is nothing special about it. It's exactly the same procedure: Wa = (Bb - Wb)/2

With white "a" in place, black playing "b" gains 2 points, so Bb = 2.
White playing "b" gains 0 points, so Wb = 0, net result Wa = (2 - 0)/2 = 1.

Having values for both Ba and Wa, we get final value V = (7 - 1)/2 = 3 (no surprise here).

What we did was recursively substituting the basic formula (B - W)/2 into itself because one of the gains (W in our example) depended on playing another move.

In principle calculation of further 'dingleberries' works the same, but this might take some practice if you want to do it in your head.
Consider V = (Ba - (Bb - (Bc - Wc)/2)/2)/2. Recursion is much easier for computers

Edit: typo corrected, thanks Robert
Edit2: some more errors detected, see below

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ChradH, do you use "gain" in a specialized meaning? If yes, which?

To mention a typo, Wa = (2 - 0) = 1 should be Wa = (2 - 0) / 2 = 1.

Allow me to try this one. JF taught us how to get a net result of +3 ( 83 - 80 ) for black in his initial diagram in his first thread from his crosses and triangles and the numbers of prisoners, taken and to be taken. How he got the crosses and triangles he is still going to explain, I hope. For the time being we neglect fractions.
Without further definition I call this number 3 the "smell" of his diagram ( that includes the number of prisoners taken ). Why smell: I dont want to reserve a more sensible name that we/JF might need later for better purposes.
I think the "gain" of a sequence Chradh is talking about is the increase in smell if the diagram ( including the prisoners and ( by using JF's still undisclosed method) the crosses and triangles ) is updated with that sequence.
Hope this answers your question, Robert.

Yes and no. The "smell" value of this position is indeed 3. Only if black really plays "a", he gains 7 solid, countable points, that's how "gain" was meant.
Of course, this is gote, so white will get compensation elsewhere. If all positions get played out, I've got an inkling that because then white will get the bigger "gain" Wx of playing the next biggest point "x", black gets By of the next biggest point "y" and so on, we might end up with nearly the same value as if we simply counted the averaged "smell" points in the first place.
But that's only speculating, let's see what JF has yet to disclose.

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