This 'n' that

jeromie · Post by **jeromie** » Wed Apr 04, 2018 1:11 pm

I don't really think the original result (observing the red dress provides confirmation that all ravens are black) is a paradox. Instead, I think it shows that humans are really, really bad at intuitively grasping the implications of very large numbers. If I have seen one black raven and was able to observe every non-black thing in the universe, then of course I would confirm that all ravens are black. The problem is that the set of non-black things in the universe is uncountably large, so I don't really learn much by any particular observation.

Your idea of defining an observational universe helps a lot with this problem. Let's say I'm sitting down to play a game of go, and my opponent hands me a bowl that is (apparently) filled with black stones. My working hypothesis is that "All go stones in this bowl are black." Making observations outside of the bowl doesn't add to my knowledge about the situation. But if I catch a glimpse of white in the bowl, only to realize that an errant piece of paper had fallen inside, it does add (some) information to support my hypothesis. Of course, you're right that it's weak evidence. I won't know for sure that there are no white stones buried in the bowl until I have examined every stone in the bowl.

The use of a statistical epistemology makes more sense in this limited universe, too. If I know there are only 180 go stones in the bowl and I have observed 179 of them to be black, that would increase my confidence in the hypothesis that all of the stones are black. Contrast this putting my hand into a bowl and pulling out a single stone at random: that doesn't tell me a whole lot about the makeup of the bowl. (Unless, of course, I allow my previous knowledge that go players tend to keep their stones separated by color to inform my hypothesis.

)

By the way, Hempel's basic hypothesis was wrong: there are (rarely) white ravens.

Also, I found a really great guide to Bayes rule and its implications. The final page on the path I followed was on Bayesian scientific virtues, and I think it sheds light on why so many people feel uncomfortable with the Carlos decision. I don't have time to outline all of that right now (and this thread probably isn't the place), but at the very least there was not testing of the advance prediction that similarity to Leela's top three moves would be highly correlated with cheating.

Bill Spight · Post by **Bill Spight** » Wed Apr 04, 2018 4:29 pm

jeromie wrote:By the way, Hempel's basic hypothesis was wrong: there are (rarely) white ravens.

Right. When I rethought Hempel's Raven in my forties, I realized that the theory of evolution should predict non-black ravens with high probability, both albino ravens and non-black varieties. A web search quickly revealed that both have been observed.

BTW, I have mentioned Bayesians who influenced me, including Savage, Good, and Keynes. The Bayesian revival of the late 20th century is interesting, because most of the people were raised on anti-Bayesian statistics, and many are unaware of the problems with Bayesianism that Keynes and others grappled with. A giant, IMO, of the Bayesian renaissance is Judea Pearl, whose writings I highly recommend. Some modern Bayesians think that the problem of the sunrise is simply solved by Cox's theorem and Bayesian updating (because that is their theory of knowledge). Earlier 20th century Bayesians knew better. Good hypothesized different levels of probability distributions, but I don't think that that really solves the problem. (Not that he claimed that it did, AFAIK.) Pearl I don't think addressed that problem explicitly, but he pointed out that certain kinds of knowledge can make other kinds of knowledge irrelevant. Then it is clear that if we know that the earth revolves on its axis, the fact that the sun rose yesterday and today is irrelevant to whether it will rise tomorrow, and that makes Bayesian updating irrelevant, as well.

moha · Post by **moha** » Thu Apr 05, 2018 6:19 am

I think most paradoxes like this are born from tricky ommissions from the problem setting, and/or any role of infinity (like two envelopes or the St. Petersburg lottery). The ravens seem no exception.

1. "all ravens are black" -> among ravens, there is no non-black
2. "all non-blacks are non-raven" -> among non-blacks, there is no raven
3. "all things are black or non-raven" -> among all things, there is no non-black raven

As a heresy I think the first form is not completely equivalent to the second. One slight difference is in worlds without ravens (makes the latter true but the former somewhat ill-defined IMO, even if theory says otherwise). And while the refutation is the same (a single non-black raven), the three forms do also carry a gradually broadening feeling (they refer the key cell from different slices of the 2x2 table, which may come into effect if the rules or the sample set of observations is not fixed).

Some information that seems missing is how many Xs are there, and how were they selected? In a world with only three instances in total, one of them not being a non-black raven is a lot of evidence, regardless of what it is. In a world with millions of ravens but only two non-black instances, a red dress is more evidence than a black raven. Even without prior knowledge of such totals, it may be important to guess whether an observed raven may have more easily been non-black, than a red dress a raven.

And just looking for the statistical probability of the existence of a non-black raven assumes no deeper logic in the assignment of labels "raven" and "black", and also that the sample set is static. But even after observing all samples the question may be extended: what if further samples were found, or what about birds born later? In this sense the minimal evidence (one less potential refutation) offered by all three cells is not the end of the story. It is knowledge of the sample generator algorithm (or the likelihoods of potential algorithms) that allows calculation of the real probability of a statement.

Bill Spight wrote:it does not matter which of the three boxes the observation falls into, the effect is the same. The observation of a black dress has the same effect as the observation of a black raven.
...
All observations of non-ravens or non-black things are very, very weakly confirmatory, if at all.

Which seem to imply that observation of a black raven is also very weakly confirmatory, if at all?

Bill Spight · Post by **Bill Spight** » Thu Apr 05, 2018 8:03 am

moha wrote:I think most paradoxes like this are born from tricky ommissions from the problem setting, and/or any role of infinity (like two envelopes or the St. Petersburg lottery).

Indeed.

For scientific purposes I think that Jaynes is right. Start with finite models and take them to the limit.

The ravens seem no exception.

Yes, since the scope of negation could be infinite.

1. "all ravens are black" -> among ravens, there is no non-black
2. "all non-blacks are non-raven" -> among non-blacks, there is no raven
3. "all things are black or non-raven" -> among all things, there is no non-black raven

As a heresy I think the first form is not completely equivalent to the second. One slight difference is in worlds without ravens (makes the latter true but the former somewhat ill-defined IMO, even if theory says otherwise). And while the refutation is the same (a single non-black raven), the three forms do also carry a gradually broadening feeling (they refer the key cell from different slices of the 2x2 table, which may come into effect if the rules or the sample set of observations is not fixed).

Some information that seems missing is how many Xs are there, and how were they selected? In a world with only three instances in total, one of them not being a non-black raven is a lot of evidence, regardless of what it is.

I address those questions by trying to define the universe of observations in such a way as to include all ravens, but to exclude as many non-ravens as possible.

In a world with millions of ravens but only two non-black instances, a red dress is more evidence than a black raven. Even without prior knowledge of such totals, it may be important to guess whether an observed raven may have more easily been non-black, than a red dress a raven.

So it would seem, and that is a popular view, among those for whom the question makes sense. Gardner suggested it, and I bought it, but eventually I came around to Hempel's view. All that matters is whether we observe a non-black raven or something else. Including a black dress.

Bill Spight wrote:it does not matter which of the three boxes the observation falls into, the effect is the same. The observation of a black dress has the same effect as the observation of a black raven.
...
All observations of non-ravens or non-black things are very, very weakly confirmatory, if at all.
Which seem to imply that observation of a black raven is also very weakly confirmatory, if at all?

You betcha!

When I calculated the posterior odds of the existence of a non-black raven, I was quite surprised to find that they were the same, regardless of whether the new observation was of a non-black non-raven, a black non-raven, or a black raven.

OC, that cannot happen if you are asking about the association between ravenhood and blackness. You have to ignore the margin totals to avoid that.

And if you ignore the margin totals, what point is a 2x2 table? You might as well have a 1x2 table, with one cell for non-black ravens and the other cell for everything else.

moha · Post by **moha** » Thu Apr 05, 2018 9:18 am

Bill Spight wrote:When I calculated the posterior odds of the existence of a non-black raven, I was quite surprised to find that they were the same, regardless of whether the new observation was of a non-black non-raven, a black non-raven, or a black raven. OC, that cannot happen if you are asking about the association between ravenhood and blackness. You have to ignore the margin totals to avoid that.

And if you ignore the margin totals, what point is a 2x2 table? You might as well have a 1x2 table, with one cell for non-black ravens and the other cell for everything else.

If the problem is narrowed down as a finite sequence of independent samples of four types A-B-C-D, then it's not surprising that the probability of the existence of a D sample depends only on the number of seen/unseen samples. But I'm not sure if this is the same as the original raven problem.

Bill Spight · Post by **Bill Spight** » Thu Apr 05, 2018 9:33 am

moha wrote:
Bill Spight wrote:When I calculated the posterior odds of the existence of a non-black raven, I was quite surprised to find that they were the same, regardless of whether the new observation was of a non-black non-raven, a black non-raven, or a black raven. OC, that cannot happen if you are asking about the association between ravenhood and blackness. You have to ignore the margin totals to avoid that.

And if you ignore the margin totals, what point is a 2x2 table? You might as well have a 1x2 table, with one cell for non-black ravens and the other cell for everything else.
If the problem is narrowed down as a finite sequence of independent samples of four types A-B-C-D, then it's not surprising that the probability of the existence of a D sample depends only on the number of seen/unseen samples. But I'm not sure if this is the same as the original raven problem.

I found Hempel's original paper online some years ago. As I recall, Hempel did not make a statistical argument, but he did say that seeing a black non-raven was confirmatory.

RobertJasiek · Post by **RobertJasiek** » Fri Apr 06, 2018 12:51 am

Click Here To Show Diagram Code: [go]$$B Initial position $$ --------------------------------------- $$ | . . . . . X . X O . . . . . . . . . . | $$ | . X . X X . . O . . . . . . . . . . . | $$ | . . . X O O O O O O . . . . . . . . . | $$ | . . . X X . . . . , . . . . . , . . . | $$ | . . . . . . . . . . . . . . . . . . . |[/go]

Click Here To Show Diagram Code: [go]$$B Locale $$ --------------------------------------- $$ | . . C C C B C B O . . . . . . . . . . | $$ | . X . X X C C O . . . . . . . . . . . | $$ | . . . X O O O O O O . . . . . . . . . | $$ | . . . X X . . . . , . . . . . , . . . | $$ | . . . . . . . . . . . . . . . . . . . |[/go]

In the Miai Values List, Sensei's Library writes that White takes a stone in sente:

Click Here To Show Diagram Code: [go]$$W Sente sequence? $$ --------------------------------------- $$ | . . . . . X 1 X O . . . . . . . . . . | $$ | . X . X X 2 . O . . . . . . . . . . . | $$ | . . . X O O O O O O . . . . . . . . . | $$ | . . . X X . . . . , . . . . . , . . . | $$ | . . . . . . . . . . . . . . . . . . . |[/go]

However, why is this sente? The answer depends on whether the following is a traversal sequence with CGT-reversal.

Click Here To Show Diagram Code: [go]$$W Traversal? $$ --------------------------------------- $$ | . . . . . X 1 X O . . . . . . . . . . | $$ | . X . X X 2 3 O . . . . . . . . . . . | $$ | . . . X O O O O O O . . . . . . . . . | $$ | . . . X X . . . . , . . . . . , . . . | $$ | . . . . . . . . . . . . . . . . . . . |[/go]

To clarify whether this is a traversal sequence, the following difference game can be considered, whose initial prisoner difference is 1 because the colour-inverse copy after move 2 is constructed by creating a white prisoner stone.

Click Here To Show Diagram Code: [go]$$B Difference game, initial prisoner difference 1 $$ ------------------------------------------- $$ | . . . . . X . X O . . . X . X O . . . . . | $$ | . X . X X . . O . . . . . X . O O O . O . | $$ | . . . X O O O O O O . X X X X X X O . . . | $$ | . . . X X . . . . , . . . . . . O O . . . | $$ | . . . . . . . . . . . . . . . . . . . . . |[/go]

To assess White's tentative traversal sequence by playing the difference game, Black starts and White must achieve at most the count 0.

Click Here To Show Diagram Code: [go]$$B Black starts I, initial prisoner difference 1, White's mistake, count 1 $$ ------------------------------------------- $$ | . . C C . X 1 X O . . . X C X O C C C . . | $$ | . X . X X 2 . O . . . . . X 3 O O O . O . | $$ | . . . X O O O O O O . X X X X X X O . . . | $$ | . . . X X . . . . , . . . . . . O O . . . | $$ | . . . . . . . . . . . . . . . . . . . . . |[/go]

Black 1 is the interesting start. White avoids the mistake 2.

Click Here To Show Diagram Code: [go]$$B Black starts II, initial prisoner difference 1, Black's mistake, count 0 $$ ------------------------------------------- $$ | . . C C . X 1 X O . . . X 3 X O C C C . . | $$ | . X . X X 4 . O . . . . . X 2 O O O . O . | $$ | . . . X O O O O O O . X X X X X X O . . . | $$ | . . . X X . . . . , . . . . . . O O . . . | $$ | . . . . . . . . . . . . . . . . . . . . . |[/go]

The mistake Black 3 gives the wrong impression that White can win the difference game.

Click Here To Show Diagram Code: [go]$$B Black starts III, initial prisoner difference 1, ko $$ ------------------------------------------- $$ | . . . . . X 1 X O . . . X 4 X O . . . . . | $$ | . X . X X 3 . O . . . . . X 2 O O O . O . | $$ | . . . X O O O O O O . X X X X X X O . . . | $$ | . . . X X . . . . , . . . . . . O O . . . | $$ | . . . . . . . . . . . . . . . . . . . . . |[/go]

If White 4 passes and lets Black 5 connect the ko, the count is 1 because of the initial prisoner difference and White loses the difference game. Therefore, White must fight the ko to possibly win the difference game. The resulting count depends on the ko and ko threats.

However, the method of playing the difference game is inapplicable if necessarily its winner depends on ko threats.

What does this mean for the tentative traversal sequence? Does an inapplicable difference game mean that there is no traversal or do we need to clarify whether we have a traversal sequence by other methods? Which other methods let us decide this? How?

Maybe White 1 - Black 2 is a sente sequence. If so, just claiming this is insufficient. We need to prove that it is either a sente sequence or a gote traversal sequence. But how?

Clarification enables determination of the correct count and move value of the local endgame in the initial position. By only pretending to have a sente sequence or else to have a traversal sequence, we do not know whether the values are correct and whether a sente sequence or long gote sequence should be played.

If the local endgame is White's local sente, the move value is 5/6. If the local endgame is White's long gote, the move value is 3/4. The counts also differ.

Bill Spight · Post by **Bill Spight** » Fri Apr 06, 2018 2:55 am

Click Here To Show Diagram Code: [go]$$Bc Locale $$ --------------------------------------- $$ | . . C C C B C B O . . . . . . . . . . | $$ | . X . X X C C O . . . . . . . . . . . | $$ | . . . X O O O O O O . . . . . . . . . | $$ | . . . X X . . . . , . . . . . , . . . | $$ | . . . . . . . . . . . . . . . . . . . |[/go]

Click Here To Show Diagram Code: [go]$$Wc Sente? $$ --------------------------------------- $$ | . . C C C X 1 B O . . . . . . . . . . | $$ | . X . X X 2 . O . . . . . . . . . . . | $$ | . . . X O O O O O O . . . . . . . . . | $$ | . . . X X . . . . , . . . . . , . . . | $$ | . . . . . . . . . . . . . . . . . . . |[/go]

After

-

the local count is 1⅔ and the temperature is ⅔.

Click Here To Show Diagram Code: [go]$$Wc Reverse? $$ --------------------------------------- $$ | . . C C C X 1 B O . . . . . . . . . . | $$ | . X . X X 2 3 O . . . . . . . . . . . | $$ | . . . X O O O O O O . . . . . . . . . | $$ | . . . X X . . . . , . . . . . , . . . | $$ | . . . . . . . . . . . . . . . . . . . |[/go]

If

-

reverses to this position, the local score is 1.

Click Here To Show Diagram Code: [go]$$Bc Black first $$ --------------------------------------- $$ | . . C C C X 1 B O . . . . . . . . . . | $$ | . X . X X . . O . . . . . . . . . . . | $$ | . . . X O O O O O O . . . . . . . . . | $$ | . . . X X . . . . , . . . . . , . . . | $$ | . . . . . . . . . . . . . . . . . . . |[/go]

After

the local count is 2½ and the temperature is ½.

If the previous diagram shows White correct play, so that the original position is gote, then the count of the original position is 1¾ and its temperature is ¾. Since ¾ > ⅔ in the previous diagrams where White plays first, the temperature has dropped after

, so White does not continue with

. There is no reverse.

----

Black komonster (territory scoring)

If Black is komonster under territory scoring, then. . . .

Click Here To Show Diagram Code: [go]$$Wc W3 elsewhere $$ --------------------------------------- $$ | . . C C C X 1 B O . . . . . . . . . . | $$ | . X . X X 2 4 O . . . . . . . . . . . | $$ | . . . X O O O O O O . . . . . . . . . | $$ | . . . X X . . . . , . . . . . , . . . | $$ | . . . . . . . . . . . . . . . . . . . |[/go]

After

the local count is 2½ and the temperature is ½, which is the same as after Black plays first. In that case we may consider

-

to reverse. And then the original count is 1¾ with a temperature of ¾, and the position is ambiguous.

----
White komonster (territory scoring)

If White is komonster at territory scoring, then the local count after

-

is 1½ and the temperature is ½. Since

is sente, the count of the original position is also 1½. With very rare exceptions somebody is komonster, so we can regard the original count as 1¾ or 1½ depending upon who it is. If we don't know who is komonster, we can regard it as 1⅔.

----

Note: Difference games are not guaranteed to work with kos, as kos destroy the independence of the positions.

RobertJasiek · Post by **RobertJasiek** » Fri Apr 06, 2018 6:16 am

Suppose we have a local endgame with a player's alternating 3-move sequence with the tentative gote traversal move value M (calculated from his traversal follower and the opponent's gote sequence follower) and the follow-up move value F in the position created by move 2. So are you saying that necessarily M > F determines 'no traversal (CGT reversal)', M = F determines 'possible traversal' and M < F determines 'traversal'?

I think difference games work well with kos if the winner of a difference game does not fight the ko but avoids all ko threats elsewhere or locally affecting the kotype by letting the opponent win and dissolve the ko. Pass if neccessary to let the opponent win the ko. It is good enough for the winner to win the difference game nevertheless.

Bill Spight · Post by **Bill Spight** » Fri Apr 06, 2018 9:15 am

RobertJasiek wrote:Suppose we have a local endgame with a player's alternating 3-move sequence with the tentative gote traversal move value M (calculated from his traversal follower and the opponent's gote sequence follower) and the follow-up move value F in the position created by move 2. So are you saying that necessarily M > F determines 'no traversal (CGT reversal)', M = F determines 'possible traversal' and M < F determines 'traversal'?

If I understand you correctly, yes. F is the local temperature and M is the original temperature, assuming 'traversal'. If M > F then why should the first player continue to play? That is inconsistent with traversal, right?

I think difference games work well with kos if the winner of a difference game does not fight the ko but avoids all ko threats elsewhere or locally affecting the kotype by letting the opponent win and dissolve the ko. Pass if neccessary to let the opponent win the ko. It is good enough for the winner to win the difference game nevertheless.

Difference games work with komonster by area scoring, and maybe with territory scoring.

Click Here To Show Diagram Code: [go]$$B Black komonster, Black has one prisoner $$ ------------------------------------------- $$ | . . . . . X 1 X O . . . X 5 X O . . . . . | $$ | . X . X X 3 4 O . . . . . X 2 O O O . O . | $$ | . . . X O O O O O O . X X X X X X O . . . | $$ | . . . X X . . . . , . . . . . . O O . . . | $$ | . . . . . . . . . . . . . . . . . . . . . |[/go]

Because Black is komonster, White cannot take the ko, but must play

. Black wins the difference game by either form of scoring.

Click Here To Show Diagram Code: [go]$$B White komonster, Black has one prisoner $$ ------------------------------------------- $$ | . . . . 5 X 1 X O . . . X 3 X O . . . . . | $$ | . X . X X 4 6 O . . . . . X 2 O O O . O . | $$ | . . . X O O O O O O . X X X X X X O . . . | $$ | . . . X X . . . . , . . . . . . O O . . . | $$ | . . . . . . . . . . . . . . . . . . . . . |[/go]

If White is komonster,

is at least as big as

, since Black cannot take the ko back. The result is jigo, which is a second player (White) win.

Click Here To Show Diagram Code: [go]$$B White komonster, Black has one prisoner $$ ------------------------------------------- $$ | . . . . . X 1 X O . . . X 4 B O . . . . . | $$ | . X . X X 3 5 O . . . . . X 2 O O O . O . | $$ | . . . X O O O O O O . X X X X X X O . . . | $$ | . . . X X . . . . , . . . . . . O O . . . | $$ | . . . . . . . . . . . . . . . . . . . . . |[/go]

fills. White wins jigo.

Bill Spight · Post by **Bill Spight** » Fri May 18, 2018 8:30 am

On combining the study of tsumego with the study of the endgame

Here, I think, is a good example. It is a modification of a problem from a Chinese book about semeai. (See viewtopic.php?p=231227#p231227 )

Click Here To Show Diagram Code: [go]$$Bc $$ | . . . . . . . . . . $$ | . . O . . . . . . . $$ | . . . O . . . . . . $$ | . . . . X . X . . . $$ | . O O O X . . . . , $$ | . O X X O X X . X . $$ | . . . X O O O X . . $$ | . . O X . . . . . . $$ +--------------------[/go]

The original problem is for Black to play and win the semeai, despite having only two dame to three. I have added a few stones on the outside to make it more well defined as a yose problem. (One defect of many yose problems is poor definition.)

Click Here To Show Diagram Code: [go]$$Bc Black wins the semeai $$ | . . . . . . . . . . $$ | . . O . . . . . . . $$ | . . . O . . . . . . $$ | . . . . X . X . . . $$ | . O O O X . . . . , $$ | C O X X W X X . X . $$ | 4 1 3 X W W W X . . $$ | 5 2 O X . . 7 C C . $$ +--------------------[/go]

@ 2

is fairly obvious, making this in a way an easy problem. However, it is easy for the solver to miss

, which is White's best reply.

I don't know if this is the main line in the book, but it should be. Note that

at 7 also wins the semeai, but

is the better play. That fact should be taken into account, combining tsumego with yose.

In the region indicated by the marked points and stones, the local territory after

is Black +11.

Since we are now studying this as a yose problem, let's look at the play when White plays first. We'll return to the play when Black plays first later.

Click Here To Show Diagram Code: [go]$$Wc White first $$ | . . . . . . . . . . $$ | . . O . . . . . . . $$ | . . . O . . . . . . $$ | . . . . X . X . . . $$ | . O O O X . . . . , $$ | C O B B O X X . X . $$ | C C 1 B O O O X . . $$ | C C O B . . 2 . C . $$ +-------------------[/go]

is the obvious play. Then after

the local territory is White +12.

OC,

is hardly a White sente, so let's look at the position after a White follow-up.

Click Here To Show Diagram Code: [go]$$Wc White first, White follow-up $$ | . . . . . . . . . . $$ | . . O . . . . . . . $$ | . . . O . . . . . . $$ | . . . . X . X . . . $$ | . O O O X . . . . , $$ | C O B B O X X . X . $$ | C C O B O O O X . . $$ | C C O B C C 3 5 6 . $$ +-------------------[/go]

The descent,

, is best. Then later

is a 1 pt. sente.

The local territory is White +15.

And that means:

Click Here To Show Diagram Code: [go]$$Wc White first $$ | . . . . . . . . . . $$ | . . O . . . . . . . $$ | . . . O . . . . . . $$ | . . . . X . X . . . $$ | . O O O X . . . . , $$ | C O B B O X X . X . $$ | C C 1 B O O O X . . $$ | C C O B C C C C C . $$ +-------------------[/go]

After

the territorial count in the marked region is White +13½.

This is a gote position. Each play gains 1½ pts.

Now let's go back and look at the play after Black plays first.

Click Here To Show Diagram Code: [go]$$Bc Black first, Black follow-up $$ | . . . . . . . . . . $$ | . . O . . . . . . . $$ | . . . O . . . . . . $$ | . . . . X . X . . . $$ | . O O O X . . . . , $$ | 4 O X X W X X . X . $$ | 3 1 C X W W W X . . $$ | C C W X C C 5 C C . $$ +-------------------[/go]

makes territory in the corner.

could also make two eyes. The result is a local score of Black +17.

-

gains 6 pts.

Backing up:

Click Here To Show Diagram Code: [go]$$Bc Black first $$ | . . . . . . . . . . $$ | . . O . . . . . . . $$ | . . . O . . . . . . $$ | . . . . X . X . . . $$ | . O O O X . . . . , $$ | C O X X W X X . X . $$ | C 1 C X W W W X . . $$ | C C W X C C C C C . $$ +-------------------[/go]

After

the territorial count is Black +11. This position is a White sente, following the main line above. The reverse sente gains 6 pts.

(No, I have not shown that it is a White sente. You are welcome to do so.

)

(BTW, you can tell that after

the local temperature has dropped because

prevents the atari at C-02 after White B-01.

)

Click Here To Show Diagram Code: [go]$$Wc Original position $$ | . . . . . . . . . . $$ | . . O . . . . . . . $$ | . . . O . . . . . . $$ | . . . . X . X . . . $$ | . O O O X . . . . , $$ | C O B B W X X . X . $$ | C C C B W W W X . . $$ | C C W B C C C C C . $$ +-------------------[/go]

In the marked region the territorial count is White +1¼. This is a gote position. Each play gains 12¼ pts.

Bill Spight · Post by **Bill Spight** » Sun May 20, 2018 2:53 am

Today we are holding a Celebration of the Life of my late wife, Winona Adkins.

At the same cemetery where we got married.

EdLee · Post by **EdLee** » Sun May 20, 2018 4:51 am

The wedding wasn't on Halloween, was it ?

Bill Spight · Post by **Bill Spight** » Sun May 20, 2018 7:10 am

EdLee wrote:The wedding wasn't on Halloween, was it ?

No such luck.

It was May 15.

goTony · Post by **goTony** » Sun May 20, 2018 1:41 pm

Bill Spight wrote:
DrStraw wrote:Once you say contrived (oops! AGA) rules I lose interest.
OK. Chinese rules.

Well played indeed!

Life In 19x19

This 'n' that

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