A page for questions and discussion about material on the Miai Values List pages.
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.)
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 "¢", 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)
4.25 / \ 3.5 5 / \ \ 4 3
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.)
4 / \ 5 3 / \ 4 25/48 / \ 3 ...
Not always. This position, for instance, is ambiguous, so that
is worth the same as
. That makes for asymmetry, since
would settle the territory.
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.
There is already a convention for trees: Black = Left, White = Right.
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.
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:
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. :-)
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 is a "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.