Sub-page of MiaiValuesList

A page for questions and discussion about material on the Miai Values List pages.

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I believe I have shown that the given values for the monkey jump can't be correct when black has no ko threats and will not get ko threats: See my page Ricky Demer's analysis regarding the monkey jump's value.

(To bring in a hopefully-familiar term: That scenario is slightly more general than "White is komaster and will remain komaster.".)

Can someone check that to see if they spot any error in my analysis?

.

===== April 28 update =====

I edited the conclusion to bring up a possible subtlety. Also, as mentioned in my previous update, I have strengthened that analysis so that it should now apply even when black has things that are already very small secondary ko threats.

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RobertJasiek: Under area scoring, how do miai values smaller than the 1/2-point button look like? - What is the explanation for "To convert from area to territory scoring we subtract 1 point, yielding -0.5." applied to the button? Under area scoring, does the button include 1 point for the stone that it occupies on the board, eh, does not occupy on the board since it is not a stone but a button?! Or must one chill any move, even a pass or taking a button, to convert to territory scoring? Why?

Bill: The basic reason that, when points in seki are not involved, the value of a play by area scoring is one point greater than the value by territory scoring is that area scoring counts the stone played, while territory scoring does not. (When points in seki are involved, area scoring may count some points that territory scoring does not.)

For consistency we apply the same principle to passes and to taking the button. So passes under equivalence scoring, such as AGA territory scoring, cost one point. If you want to consider taking the button as a pass, then you get 0.5 point for taking the button, but lose 1 point for passing. Net result: -0.5.

Taking a dame dominates taking the button. It would not do that if taking the button were worth more than it was.

RobertJasiek: What exactly does "for consistency" mean here?

Bill: By area scoring playing a Japanese dame is worth 1 point. By territory scoring it is worth 0. By area scoring taking the button is worth 0.5 point, 0.5 point less than playing a dame. For the values to be consistent, taking the button by area scoring should also be worth 0.5 point less than playing a dame, and therefore worth -0.5. Any other score would be inconsistent between scoring methods.

Now, we don't normally use a button with territory scoring, but let's pretend that we did, say, with AGA rules. Say that Black plays the last dame, then White takes the button, and then there are two play ending passes. Say also that the territory difference on the board is 6 points in favor of Black. The area difference is 7 points. Since White took the button, the area difference before komi is 6.5 points. If White loses 0.5 point for taking the button under territory scoring, the territory difference before komi is also 6.5 points. That's consistency. :-)

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i am not sure i get the miai value, but i am not sure i get the deiri value either... what is the deiri scoring here ? i find a 1 point sente for black or a 2 point gote, followed by another 2 point gote. what do you think ? is it correct ? and can we easily convert from deiri to miai (i find deiri easier even though not as powerfull)

blubb: Beside the obvious drawback that Deiri doesn`t take into account the number of additional moves, there is another, maybe even more serious flaw: based on Deiri alone, you can`t tell where in the tree sente ends. On a quick glance at this case, based on the normal value tree and the conversion miaiV = 2 * normV2 - 1 (where normV2 is the second number of the normal value), I get a miai value of 1.00 here. If I am wrong, please, let me know.

iago: thanks for your reply, the thing is, with miai value i dont quite get the sente / gote part of the move... is it 0.92 sente for black and 0.92 gote for white ? or is it just irrelevent ? with deiri counting i feel that there are two different count based on how white want to treat the play here...

Bill: It is a 1 point ambiguous move.

iago: Ambiguous move was the page I was looking for indeed... but then, why is the capture 3 stone (1.92) not ambiguous ? see the example below

PJT 2019-02-10: Articles

Ambiguous Move&Ambiguous positionhave since been distinguished.

Bill: It is gote. Capturing 3 stones, even when leaving a stone in atari, is bigger than capturing that stone back.

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Calvin: Shouldn't an example of the first line double gote hane be included here, like the first example in Miai Counting Made Easy?

Bill: Oversight corrected. Thanks, Calvin. :-)

Calvin: I don't see how this is the same value as the previous position, unless you assume *a* is white's sente next; but that just creates a 1/3 point end position.

cynewulf: If White plays *a* when the ambient temperature is above 1/3, then Black will ignore and play a larger move. Otherwise, Black will connect, or play a move of equivalent size.

Calvin: Thanks. If I understand correctly, in essense after , white *a* would then be a zero-point move.

Calvin: So could this be included in the 0.0 section? If the definitions here assume (and I think they do) that the shown moves are the biggest moves in the game, that means that if white is playing , that means there is nothing better for black than to connect this or play another move of the same size.

and .

So we can consider this position to have miai value of -1 (the same as a pair of dame). We could also consider it to be a 0 point sente for White. ;-)

1.06

Can somepne verify the enter a small room value?

Black move leaves white nothing since a move which creates at 1/2 point maia move is itself worthless since it is a mistake unless all fake half point kos are filled in which case it is 0 points sente. (Much like making the first third of a ko is worthless)

White defend and now white can play an additional stone to get 3 points or black can play leaving white 2 points and black a 1/2 point move. Thus if white defends he gets 2.25 points.

The maia value should be 1.125.

Bill: Here is the game tree.

A / \ 0 B / \ C -3 / \ D -2 / \ 0 -1

D is worth -0.5. C is worth -1.25. B is worth -2.125. A is worth -1.0625, and so the player who moves from A gains 1.0625.

I see now. The move black gets is not a half point move but rather a corridor crawl worth .75. Even this unlabled game tree is very informative. Perhaps it would be good to show the tree next to each position. It would certainly help people see where the maia score comes from as well as efficiently giving the score of dozens of related positions.

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KarlKnechtel: I did a bunch of math with pen and paper just now, and got a miai value of 1.98 for this situation (2 - 1/48) and 1.04 for the two-stone capture. But I don't think I have a perfect understanding of these calculations, and also it's 3:30 AM here. If it will help global understanding any, I could explain my calculation later on but I don't suppose that should go on this page :\

Andre Engels: My calculation gets to 1 11/12, which is indeed 1.92 rounded to 2 decimals. Here is the diagram:

1 11/12 / \ / \ 3 5/6 0 : X O O O O O, no prisoners / \ / \ 5 \ X . . X X O, 3pr. \ \ 2 2/3 / \ / \ 3 1/3 2 X . X O . O 2pr. X . O O . O, 2 pr.

Note: the 1 11/12 is not the 1 11/12 in this diagram, but the difference between the 1 11/12 and the 3 5/6.

Karl Knechtel: I have the **2** case in my figuring split up further, reflecting the fact that if black makes the next two plays from that position, he captures again. And we allow black to move twice to find the **5** case, so I don't see how it's different...

Anyway, my values are 1/3 for a single stone (simple ko), 25/24 for two stones, 2 - 1/48 for three stones, and in general (n - (13/12) + 2^(-n-1)) for n>=3 stones. The above derivation is taking out the little exponential correcion for the n=3 case.

Bill: Karl, what do you mean by the **2** case and the **5** case? Do you mean the scores? They are the local scores. That's it. There is no further move in the 5 case. In the 2 case Black has a ko threat, that's all. Another way to look at it is as a miai. White fills and Black plays the dame, or **vice versa**, except that if Black plays first it's a ko threat.

Andre Engels: The difference is sente or gote. In general, we are looking at diagrams of this form:

??? / \ / \ (A+B)/2 C / \ / \ A B

In general, the value of ??? will be 0.5C+0.25(A+B). But not always. If A is much larger than B, then the first player's move to (A+B)/2 will always be answered by a move to B. In that case, provided B is greater than C, the first player will move to (A+B)/2 at some time, and the second player will answer by moving to B, so the actual value of A and C does not matter, and the value of ??? is B. In general, it is the lowest of the two values.

Tamsin: Could somebody clear up one thing for me please? I note many calculations produce numbers with decimals (what's the mathematical term for such numbers?), e.g., 2.3 or 1.93. Now, given that the smallest unit in a go game is one point (discounting the half-point in komi), does it matter if one plays a move evaluated at, say, 1.52 instead of one evaluated at 1.94? (All other things, such as sente/gote considerations, being equal.)

HolIgor: The unit is half point. So, as Bill claims it is enough to have values like 1.5+ and 2.0-. In the end of the game the decimal points are converted into the whole ones by getting the last move. I don't know if 1.98 has more chances to become a point in the end of the game than 1.92. Perhaps this can be proven, perhaps not.

Bill: See Numbers for an example where a mistake of 1/32 point ends up costing a point. ;-)

To expand upon Andre's point, say that we have this kind of situation:

A / \ / \ / \ B C / \ / \ D E F G

where D >= E >= F >= G.

The idea is that E >= A >= F, because if Black plays first White can reply and hold Black to E, while if White plays first Black can reply and hold White to F. (This does not mean that the eventual local result won't be D or G. But in that case, with correct play there will be compensation elsewhere.)

*Ok. So we should evaluate A as max(F, min (E, (D+E+F+G)/4))? I.e. the normal value is given by the usual averaging, but must be at least F (black response to white's first play being sente) and at most E (white response limiting black sente). Did I get that right? -- Karl Knechtel*

Bill: Right, Karl. :-) (Subject to the restraints, OC.)

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When referring to "points", there seems to be some confusion between miai and deiri points. Since we agreed to go to 2 decimals, I propose to measure points from miai counting in "cents" and write it as e.g. "33 ¢" or "450 ¢". The first occurrence on a page could be written as "¢" -- Sebastian

Deebster: Could we use a character everyone has on their keyboard?

Sebastian: Actually, entering it is not a problem - on any keyboard, you can type "**&****#162;**", and it will display nicely as "**¢**".

But just because I love choices: Another alternative would be to spell it out - similar to what Europeans do if they don't have the Euro sign.

Bill: What confusion? Both types of values measure differences between the counts of different positions, which are expressed in points. Differences between points are also points.

Robert Pauli: My **prefered format** would be

(Hope, example is OK: 'a' worth 0.75 - W's 2nd thrust is sente, suppressing branching)

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4.25 / \ 3.5 5 / \ \ 4 3

- 'a' instead of or - there is symmetry
- black territory - no negatives
- White goes
*left*(< 0), in diagrams*and*trees - tree, shows computation and sente (without explaining sente, however), and additionally makes clear distinction between MV's of positions and moves
- besides headings, exact figures to enable checking (maybe 3/4 instead 0.75, certainly 1/16 instead 0.0625)
- "1 + 11/12" instead "1 11/12" (or are there "hard blanks" ?)

Charles Not everyone can compare fractions in their heads, though.

ilan: The correct notation would use fractions, but with integers written in binary, as most of the rational numbers appearing have denominators powers of 2 (just like English length measure).

Charles Except in cases involving *ko*. OTOH small kos are common in the endgame. I actually *like* the compromise of using *cents*; if I were to study the endgame intensively, I think using a scale from 1 (well, 33) to 1000 cents would be quite user-friendly. Of course theorists will want exact values; but this seems to have not much practical value.

ilan: Like I was saying, if you're a British or American carpenter, go endgame arithmetic should not be a challenge. Otherwise, use fraction or floating point arithmetic, but in base 2.! 0.75

Bill:

(* shows Robert's comments.)

(Correcting example and tree.)

Not always. This position, for instance, is ambiguous, so that is worth the same as . That makes for asymmetry, since would settle the territory.

PJT: Is it an ambiguous position because is an ambiguous move?

- black territory - no negatives

These pages were set up with Black to play. (That's not always followed, I guess.) Anyway, negatives are hard to avoid without padding the diagrams with independent Black territory.

- White goes
*left*(< 0), in diagrams*and*trees

There is already a convention for trees: Black = Left, White = Right.

- tree, shows computation and sente (without explaining sente, however), and additionally makes clear distinction between MV's of positions and moves

I think that that's too much to show. It's only a list. However, we have better linkage on SL, now, so providing links to the relevant followers should give the same information.

Later: I see that most of the diagrams lack count information, which is necessary for making calculations. I have started adding count info.

- besides headings, exact figures to enable checking (maybe 3/4 instead 0.75, certainly 1/16 instead 0.0625)

I like exact fractions, myself, but decimals are good enough. Links would enable checking.

Robert Pauli: Outch, 3.5 in my tree was wrong. Since a sente follow-up doesn't change a position's MV, I should have copied the 3 up:

4 / \ 3 5 / \ \ 4 3

That White's second thrust *is* sente, more or less is intuition, but one can also check it via the method of multiples (4 * 3 > 4 + 4 + 1 + 0).

Bill, you seem have done the complete tree - let me try (and maybe have others understand 25/48):

4 /\ / \ 3 5 /\ / \ 25/48 4 /\ / \ -47/24 3 /\ / \ -53/12 1/2 /\ /\ / \ 0 1 / \ -6 -17/6 /\ / \ -4 -5/3 /\ / \ -7/3 -1

Normally parent nodes average their childs, however, they must stay between grandchilds A and B:

P / \ / \ /\ /\ A B

A <= P <= B

That's why it's **3** under 4, and not 2 + 1/4 + 1/96.

Decimals of course are OK, but rounded ones hamper checking. Therefore at least each rounded one should be accompanied by an exact fraction.

It might not always be symmetric (example please), but if, it's better as I said.

Since (number) trees in ONAG increase going right, why not our's too?

Bill: IIRC, these trees are like the ONAG trees, rotated by 180 degrees. Anyway, it's the same people. See *Winning Ways*, *Mathematical Go*, *Games of No Chance*, etc.

Anyway, I get the same values you do. :-)

Robert Pauli: They're not rotated, they're mirrored :-)

Bill: Well, then, the numbers increase to the left in ONAG, too. Oh! do you mean trees **of** numbers? In those the numbers increase going right in our trees, too.

Example:

The regular game tree of this position looks like this:

0.5 / \ 1 0

Where 0.5 is the count.

But in chilled go this is a number, and its game tree looks like this:

0.5 / \ 0 1

Chilled go aside, nearly all numbers in go are integers.

Robert Pauli: Bill, somehow you're making *my* point, not? :-)

Bill: I don't think so. I think you want trees for both numbers and **hot** games to have values increasing to the right. No?

Robert Pauli: Yes - why have it this way *and* that way?

Bill: Because {1 | 0} and {0 | 1} are two different games. The second one is a number, 0.5. The first one has a mean value of 0.5, but is not a number.

Robert Pauli: Replaying a game, there was this:

I concluded at to be sente, but , taking back, not . . . making worth 5/12 (from 5 + 7/12 to 6). Right?

Now, if yes, add two 1/2 opportunities (Black can make an eye or White can prevent it). Three options in all.

Obviously Black should start with above, even if its miai value is 1/12 less. This gets him 7, compared to 6 + maybe 1 when starting with an eye (and at . . .) - but what use are miai values then ??

Bill: Nice position. :-) It's a 2/3 point sente.

Robert Pauli: Only if *is* sente (actually I first also had +2/3 :-), however, then I took 4 copies of the position after , White starts:

- if Black always answers, it's 4 x (1 + 1/3) = 5 + 1/3
- but if not, it's 3 + 3 + (-1) + (1 + 1/3) =
**6**+ 1/3

Bill: You are right that is not sente. However, that is not required. only has to be hotter than .

has a miai value of 2/3. has a miai value of 2 19/24. has a miai value of 1 5/12. has a miai value of 1 1/6.

As long as White starts play at an ambient temperature of less than 1 1/6, Black should reply. Between that temperature and 2/3, White has the privilege of playing with sente *in the original position*.

Try it with 4 copies of the original position and have White play before switching to another copy to play .

Robert Pauli: Thanks, Bill. Reproduced your figures and agree: 2/3. What did I learn? Well, it seems that miai values aren't context-free, that's, if during analysis of some position another shows up, I just can't take an already made analysis of the second, but rather have to re-evalute its sente clipping in the context of the surrounding tree. Bad news to me. :-)

Bill: De nada. :-)

But it's not such bad news. You can still use information from your previous analysis. First, you calculated that was a 5/12 point sente. But you already knew that the local temperature after is 1 5/12. So White should play again.

Next, assuming that - is gote, you find that it has a miai value of 11/12. But the local temperature is now 1 1/6. Black should play again.

Finally you calculate a 2/3 point sente, with a final position having a local temperature of only 1/3. Fini. :-)

Robert Pauli: You mean I "simply" could have started with pretending every move to be gote (Black goes right, sorry)

(4 19/48) - /\ | / \ | 1 29/48 / \ | (2 19/24) 6 - /\ | / \ | 2 19/24 / \ | 0 (5 7/12) - /\ | / \ | 1 5/12 / \ | (4 1/6) 7 - /\ | / \ | 1 1/6 / \ | 3 5 1/3 - : | : | 1/3 | -

recognize temperature increasing at move 2 and cut off the outer one (towards 0)

(5 7/12) - /\ | / \ | 5/12 / \ | (5 2/12) 6 - \ | \ | 5/12 \ | (5 7/12) - /\ | / \ | 1 5/12 / \ | (4 1/6) 7 - /\ | / \ | 1 1/6 / \ | 3 5 1/3 - : | : | 1/3 | -

again recognize temperature increasing, at move 3 now, and again cut off the outer one (towards 7)

(5 1/12) - /\ | / \ | 11/12 / \ | (4 1/6) 6 - \ | \ | 11/12 \ | (5 1/12) - / | / | 11/12 / | (4 1/6) - /\ | / \ | 1 1/6 / \ | 3 5 1/3 - : | : | 1/3 | -

once again recognize temperature increasing, at move 4, and once again cut off the outer one (towards 3)

(5 1/3) - /\ | / \ | 2/3 / \ | (4 2/3) 6 - \ | \ | 2/3 \ | (5 1/3) - / | / | 2/3 / | (4 2/3) - \ | \ | 2/3 \ | 5 1/3 - : | : | 1/3 | -

and finally be satisfied with a non-increasing temperature? That wouldn't be that bad (and would do away with the copies). Hope embedding such subtrees doesn't undo cuts made in them . . .

Bill: Only the correct figures will not be inconsistent. And fortunately, when they are wrong, they are systematically wrong. ;-) You can make any assumption about sente and gote and test it.

One point, though. Except when you start at temperature 0 or less, you should not be satisfied with keeping the same temperature, but continue.

Oh, and looking at your method, I think you may have to backtrack if you work top down.

For instance, suppose that you start with this tree.

1.3125 (1.3125) / \ (1.375) 2.625 0 / \ 4 1.25 (2.25) / \ (1.5) 3.5 -1 / \ 5 2

I have filled in means and temperatures (in parentheses) on the assumption that each move is gote. Using your method, I think you generate this tree:

1.75 (1.75) / \ (1.5) 3.5 0 \ 2 (1.5) / (1.5) 3.5 / \ 5 2

Since the temperature drops at move 2, you should backtrack and restore the play to 4. But that is not so obvious, I think.

I find it easier just to erase the means and temperatures and retain the scores, at least at first, until we get a consistent subtree. Then I think it's OK to prune.

1.3125 (1.3125) / \ (1.375) 2.625 0 / \ 4 1.25 (2.25) / \ (1.5) 3.5 -1 / \ 5 2

We see the top inconsistency, and erase the top mean and temperature.

A / \ (1.375) 2.625 0 / \ 4 1.25 (2.25) / \ (1.5) 3.5 -1 / \ 5 2

There is another inconsistency at the next level. Let's erase the next mean and temperature.

A / \ B 0 / \ 4 1.25 (2.25) / \ (1.5) 3.5 -1 / \ 5 2

There is no inconsistency lower down, so replace B. (We can prune at C).

A / \ (0.5) 3.5 0 / \ 4 C / (1.5) 3.5 / \ 5 2

Pruning the first "external" branch, we try a mean of 3.5, but that makes a temperature of only 0.5, which is inconsistent.

A / \ (1) 3 0 / \ 4 C / D \ 2

Next we go one step further and find a mean of (4 + 2)/2 = 3. That makes a temperature of only 1, which is consistent.

Now we replace A.

1.5 (1.5) / \ (1) 3 0 / \ 4 C / D \ 2

This is consistent, and we are done, except that we can simplify the tree.

1.5 (1.5) / \ (1) 3 0 / \ 4 2

BTW, I think our discussion rates its own page about method. :-)

Emeraldemon: I'll try: MiaiCountingWithTrees

Robert Pauli: Thanks for your counter-example, Bill. I should never feel sure! I took my time, hoping to creat smart questions, but here I am:

My method indeed would create your second tree, even if I chose temperature 1.75 instead 1.5 for moves 2 and 3. Any difference?

Bill: I am not sure what you are asking. If you prune the move to 4, you have to backtrack later and restore it, anyway.

RP: I meant that using my (wrong) method I had (or chose to have)

1.75 (1.75) / \ (1.75) 3.5 0 not (1.5) 3.5 \ 1.75 (1.75) not 2 (1.5) / (1.5) 3.5 / \ 5 2

Let's now assume that no cut was done. Which tree would be the correct one - yours, mine, or one "between" ?

Delaying cuts (at A and B) sounds good, but why "can (we) prune at C" ? This was a consistent subtree (rooting in C), and you (I'm pointing with the finger ;--) cut it ?

Bill: Well, actually, it has to be consistent before we can prune it. Only if it is consistent can we be sure that the values are correct.

We can prune White's follower from C because we already know that our assumption that White's play to C is gote produces an inconsistency. Therefore Black should reply and White never gets the chance to play from C.

BTW, pruning is not necessary. The full tree, with means and miai values, is

1.5 (1.5) / \ (1) 3 0 / \ 4 1.25 (2.25) / \ (1.5) 3.5 -1 / \ 5 2

You have to examine it a bit to see that the mean value of 3 and the miai value of 1 are correct. Pruning makes that easier to see.

RP: Hmm, rather stick to pruning . . .

It didn't work (0.5 < 1.5), and you cut it again (cutting off 5). Is it cutting off a subtrees "external" branches until it creates no inconsistencies above?

But why is one sure that, after having trimmed the subtree, one couldn't undo one cut in it and still be consistent ?

It seems that not only do we want the tree to be consistent (temperatures dropping on the way down), but as well bring its temperatures down as low as possible. Right ?

Bill: Are you talking about keeping going if you get the same temperature (greater than zero)? You need that information to determine sente, gote, and orthodox play. Anyway, if there are no kos, there is only one tree with correct values.

RP: I meant that my false tree was also consistent in that its temperature never increased: 1.75 >= 1.75 >= 1.75 > 1.5. If that's not enough, there has to be a second goal. Your result gave 1.5 > 1 >= 1 >= 1, which are lower figures. But it might also be that the second goal just is that no cut can be undone without creating an inconsitency, which my tree misses.

Is trying out, a "method" ? (Sorry, ugly one ;--)

Bill: Well, yes. If you make a guess about sente and gote and it checks out, you are done. That is essentially how the pros calculate during a game, except that they do not always check fully and sometimes make mistakes. (Actually, my guess is that they rarely check, trusting their judgement.)

RP: Like normal endgame books. They just tell you that, hey, this

isa "sente" follow-up - believe it or not.

Bill: Let's look at your overpruned tree.

1.75 (1.75) / \ (1.75) 3.5 0 not (1.5) 3.5 \ 1.75 (1.75) not 2 (1.5) <-- / (1.5) 3.5 / \ 5 2

Where does the mean value of 1.75 (indicated by the arrow) come from? These values do not percolate down, but up. The top values are derived from those below, not the other way around. Pruning White's follower is like treating the mean value there as <= 2. The tree must be consistent with *all* such values. 2 is the most problematic value, and that's the one to use. That drops the temperature to 1.5 after Black's first play, which is inconsistent with the first pruning, so we need to restore Black's follower(s) from that point.

Robert Pauli: Hmm, you said that "pruning White's follower is like treating the mean value there as <= 2", but doesn't the mean at X

: X / / / / (1.5) 3.5 / \ 5 2 |--|

*always* has to be <= 2 because of this "stay between descendants" rule - independently of pruning ?

Bill: Well, yes. But what I was focusing on was that pruning ignores the information from the White follower of X, so that X can be any value up to 2, and should be consistent within those limits. Before pruning we had a specific value. After pruning we do not. X is not constrained from above, but it is from below.

Robert Pauli: Abe claims in Dramatic Moments on the Go Board (p. 68) that

is worth 2 + 11/12 = 3 - 1/12.

But when I (naively) drew the tree (dots deduced)

3 - 1/12 . - / \ | 3 - 1/12 6 - 1/6 . 0 - / \ | 1 + 1/6 7 . 5 - 1/3 - / \ | 2/3 5 + 1/3 . 4 - / \ | 1 5 . 3 - / \ | 1 6 . 4 - / \ | 1 7 5 -

I recognized the temperature increasing from 2/3 to 1.

Since, Abe can't be wrong (grin), I checked with Method of Multiples, taking eight copies and letting Black start:

4 x 4 + 4 x 0 + 2 x 3 - 2 x 1 + 2 1/3 + 1 = 23 1/3 = 8 x (3 - 1/12).

So, Abe sensei seems to be right -- but what about the temperature? Didn't you tell me somewhere that it may not increase, Bill?

Bill: We have to stop at a stable follower. The increase in temperature is temporary. In the case of sente, it lasts until the final reply. In the case of reversal, as here, it lasts until the final gote. So we can simplify the tree like this:

3 - 1/12 . - / \ | 3 - 1/12 6 - 1/6 . 0 - / \ | 1 + 1/6 7 . 5 - 1/3 - / \ | 2/3 5 + 1/3 . / . \ 4

A common reversal is the hanetsugi.

Robert Pauli: Thanks, Bill.

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Is it completely correct to say that you should add one for area scoring, except for some sekis and kos. (Also, territory scoring is not monolithic here. Some versions count territory in seki, for instance.)

I thought a dame had a value of a half, for example.

Bill: Nope. If Black gets a dame, the local area score is +1, if White gets it, it is -1: {1 | -1}.

Also, I think I've heard of ko fights for the last dame in chinese scoring, but I might have got the wrong end of the stick.

Bill: I think what you have heard is about one player's filling or taking and filling the ko while the opponent has no dame. If the territory miai value of the ko is 1/3, then its area miai value is 1 1/3. Since the opponent has no play to offset filling or taking the ko, except to fill territory, the player who wins the ko may gain 1 or 2 points more by area scoring than by territory scoring. See komonster.

If there is anything interesting to say about miai values for chinese scoring, I'd be keen on having a page for it.

1/9 = ~0

The reference result is copy three, where white has one point. He has two points in copy one and zero in copy 2, total of 3 points. There is one white ko stone in atari left in copy two, which is worth 1/3 points for black. Black has thus gained 1/3 of a point is sente in a total of three copies, which makes 1/9 of a point. However, if you increase the number of the copies, black cannot increase his gain above 1/3, because instead of creating another instance of copy three, white may connect a ko at 5 in a previous instance of copy two. -- MattiSiivola

Bill: First, let's look at this position with different komasters.

When White is komaster White will get one point of territory, which by convention we write as -1. (Under area scoring there are positions where White to play might throw in and win the resulting ko after all the dame have been filled. But then White would be komonster.)

When Black is komaster and throws in, there are two possible swings.

In this case there is a swing of 1 point in 2 net plays (one net play for Black when he plays first, one play for White when she plays first), for an average gain of 0.5 per play.

In this case there is a swing of 2 points in 5 net plays, for an average gain of 0.4 per play.

When Black is komaster the gain from the throw-in lies between those two values. It is 4/9 on average, and the original count in the corner is -5/9.

What about when neither player is komaster? We know that the original count lies between -1 and -5/9, inclusive, but there is no single theory that tells us what it is, and hence, no single theory that tells us what the throw-in gains.

Under a neutral threat environment (NTE) the count is -6/7 and the throw-in gains 1/7. But that is unrealistic. Here is one possible line of play under the NTE.

plays threat. fills the ko. completes the threat.

White has 2 points, while Black has gained 1 2/7 points from the ko threat, for a net "local" score of -5/7.

But, OC, a score of -5/7 is impossible. And Black's ko threat is impossible.

The NTE theory provides only an approximation to reality; here the error may be 2/7 point. :)

Matti Siivola is taking a different approach. But there is a problem with copy 2. The ko is left unfinished. But then the local temperature is 1/3, which is greater than 1/9, so play has not reached a stable state. Play needs to continue.

A miai value of 1/9 suggests that if there were 9 copies of the position, White would get 8 points with correct play, regardless of who plays first. That may be so. If so, it would provide a basis for the 1/9 miai value. :)

Bill: Using the method of multiples I think that there is an argument for a miai value of 1/6. :)

After 14 moves, White has 5 net points in 6 copies, regardless of who plays first. The average count is -5/6, for an average gain per play of 1/6.

Matti: Someone wrote at the main page that if black is KoMaster then his move is worth 4/9 points. Master for which stage of the ko?

Bill: Komaster for the 7/9 stage. Komaster for the 1/3 stage does not matter, as 1/3 < 4/9. Here are the relevant sequences of play when Black is komaster.

elsewhere. There may be ko fight between and .

Black moves to a position worth 1/3. There is a swing of 4/3 in 3 net moves, for an average gain per move of 4/9.

, , elsewhere. fills at .

If Black is komonster he may be able to win both kos. Note, however, that there is a swing of 2 points in 5 net moves, for an average gain of 2/5 per move. 2/5 < 4/9, so Black has lowered the temperature of the ko fight, which requires being komonster. It is possible, in that case, that this sequence may occur with best play. :)

If Black is komonster it is also possible, maybe more probable, that White will simply play instead of allowing Black to take and win the second leg of the ko.

Matti: It seems that possible dedominators for values of moves are powers of 2 and powers of 3, possibly multiplied together. 5 is also possible.

Matti: 0,39 (7/18)

Bill: When White is komaster the play is in the indicated ko, with a gain of 1/3. When Black is komaster the play is in the throw-in ko, with a gain of 1/2.

Matti: For clarity I added a black stone. Let's first have thee copies.

In copies 2 and 3 white has 2 5/6 points. Now, let's assume that we had 18 copies of the original position and have now 6 positions of copy one.

As shown above white has 5 points in six positions in the corner. In one position of the six he has one point at the edge and in five others two points at the edge, total of 16 points in six positions, thus 2 2/3 points. So black has gained 1 + 2 5/6 - 2 2/3 = 7/6 from total of three kos. One ko is thus worth 7/18 points.

[edit]

RobertJasiek: The text says that the next play (after move 1) had the value 4/3. I think it has the value 1 5/12. As a consequence, move 1 has the value 1 19/24. Can somebody please confirm?

Bill: OK.

After , **a** and **b** are miai. That is, if White plays at **b** Black's reply is at **a**, and if Black plays at **a** White's reply is at **b**.

RobertJasiek: Why miai? This conjecture should be supported by a value analysis.

Bill: Consider the play at or above temperature ⅓. The mast value (count) at that temperature lies between the result after Black

a- Whiteband the result after Whiteb- Blacka. These positions are the same and their mast values are equal. Since there is only a ⅓ pt. ko in that position, the mast value is easy to determine. See diagrams below.

RobertJasiek: More analysis is necessary to decide whether Black a - White tenuki or White a - Black blocks - White tenuki are better.

Bill: Black

agains ⅓ pt., leaving a sente ko because if Black takes the ko it is sente. After that, Black can fill the ko, OC. The second line leaves a position with the same count and miai value. Black to play can take two stones, White takes back, and Black plays ata, making a ko, just as if Black played atato start with. White to play connects atb, then Black takes two stones and White takes back, leaving the same score as if White connects atb, Black plays ata, and White fills the ko.