wms's Dame-Seki Challenge
wms: I have a problem, but I won't put it as problem of the month because it is not a "typical" go problem.
Lately in rec.games.go a discussion has sprouted up and KGS' implementation of Japanese rules. That jogged my memory about something I thought about a while back, but was never able to prove one way or the other. It occurred to me that it might make a nice problem for people interested in go oddities.
The problem comes from KGS' seki detection code. The goal of the KGS seki detection code is to always correctly spot seki, as long as both a) all dame that can be filled, are filled, and b) all dead stones are marked correctly.
But I thought of one situation that would make the KGS algorithm have an error. I think it is not possible for such a situation to exist, but I'll put it here, and invite people to try to find such a situation, or prove that it cannot exist:
Can you find a position where one player (black, let's say) has two one-eyed seki groups separated by a bamboo joint (or similar connected-by-dame-but-can't-connect shape) of white's. If black tries to push through the bamboo joint, black will die, but if white tries to fill one of the "gap" dame in the bamboo joint, then white will suffer some problem (either dying or allowing one or more black seki groups to become fully alive).
If anybody can find such a board position, then it will be interesting, and I will have to adjust my seki algorithm. If somebody can prove it can't exist, then that's great, my algorithm is fine.
If you want to know more about the algorithm itself, just use google to search rec.games.go for the terms "shubert" and "seki".
rubilia: Here's the direct link.
wms: OK, to help things along, here is an example position that is close but not quite there:
This looks like what I need, but note that if black plays first, then black can connect, make life, and kill white by pushing through the bamboo joint; if white moves first, white can just fill the bamboo joint and end up killing both black groups. So this isn't even a seki at all, but it shows what I need - two one-eyed groups, separated by dame. I need a position like this, but where neither player can fill the dame separating the two one-eyed groups. Or, just as good, I need proof that in such a situation, one player will always be able to fill the dame.
impu1se: I think something like the second example under number 4 in StrangeSekis is what you're looking for. wms: No, #4 in strange sekis doesn't match - the two one-eyed groups are opponsing colors, so "connecting" them won't give a living group.
Like this? Neither can attempt to make a move inside the bamboo
Because if you want them to have eyes it's clearly impossible as filling the dame and connecting would saperate the other 2 killing them... <- That's if I understand you correctly. Because 2 eyes always win vs anything else... And an eye wins vs none providing there aren't enough outside liberties - But if there it wouldn't make sense in the first place...
wms: The example above doesn't work, KGS/CGoban scores it perfectly - there is only a single one-eyed group. Again, look at my example above, the key thing is two one-eyed groups of the same color, connected by dame that neither player can fill. You say that it's clearly impossible, but I don't see that; my point is that filling the dame must be impossible. Don't forget, when you fill dame, you also under certain circumstances reduce your liberties - so there are cases where dame simply cannot be filled safely.
firefox04?: Wms, in your example, what if B attempts to capture the 2 stones?
- wms: Then white dies. :-) That's why my example isn't a solution, but it's just an attempt to show what I am looking for.
(Sebastian:) How about awarding an annual KGSPlus membership to the winner?
- wms: Annual? Well, no, but I will give a full 3 month membership to anybody who finds such a position. So try to find one!
Scatee?: Like this black has two eyes and neither black nor white can file those dames.
wms: Oh, very very close! The only issue here is that the two black one-eyed groups must be able to reach each other through the seki spaces. But this really is very close.
Bill: wms, I think this meets your criteria. :-)
yoyoma: Can't white choose to kill either the two top groups or the two left groups here? When he starts to kill the left groups, the top ones can connect and live, but that's still better for White than leaving it as seki right? Same comment for Position 2.
Bill: Well, by Japanese rules, if these positions are left as is at the end of the game, they are seki. In addition, this one has a temperature of -1, the same as seki.
yoyoma: But wms is looking for situations where neither side wants to add a move. Here White wants to add a move to kill the two left groups, instead of leaving them as seki.
Bill: I think this one does, too, but I like the first one better. ;-)
seldon: But both players will very much want to play there, white to kill all, black to save at least one group:
seldon: Note that if black 2 at 3, white a. black can't make two eyes with the lower group, and white has two liberties to win the semeai.
seldon: If black goes first, white is short one liberty to kill both, so one black group (the lower, since it's worth less) will live.
Bill: OK, here is position where neither player wants to play first by territory scoring, since it makes a difference of one point. (To prevent seki, White will have to play first before ending the game.)
wms: Very interesting, Bill. But it seems to me that by territory scoring, it is better for White to play, killing two of the black groups, letting the other two, live, right? If white does nothing, then nobody has any points. If white kills two groups, then white gets 22 points (kills 10 stones, gets 12 territory) and black gets 2. So really this is just an unfinished position, White needs to take the last 20 points profit, right? Yoyoma had my intent correct - sure, you can construct a position where if players refuse to move, then you get a weird seki that CGoban will score incorrectly. But I'm looking for a position where all moves with nonnegative value have been played, but cgoban still scores incorrectly.
Bill: In this position neither player wants to play in the same sense that neither player wants to play in three points without capturing. Letting the other player play first is worth a point. By Japanese scoring, White needs to play here to avoid seki, but that is an artifact of the rules. By other territory rules, such as Lasker-Maas or mine, the players could agree to the score at the end of the game or play the position out in the encore. (I do not think it would be played out under the Japanese '49 rules, either.) All moves with non-negative values have already been played.
These examples were based upon your suggestion that White's killing the two one-eyed Black groups might allow another Black group to become fully alive. IOW, if White has one less liberty that group can become fully alive. But Black can enforce that (or full life for the disconnected groups) himself by playing a dame which threatens to connect. If Black does that, it may cost a point for losing the played stone, by comparison with White's killing the two groups. Such a position is going to be like these, where by Japanese scoring White does better to kill at the very end of the game than to allow seki.
For White to be able to kill when Black attempts to connect, White needs two extra dame (possibly including a dame in an eye or false eye). Therefore killing immediately leaves these White stones with at least three dame. These stones cannot be killed. If reducing the dame to three allows other White stones to be killed to offset the loss of the Black stones, then there seem to be two cases.
1) They must be disconnected from these stones and if Black fills one of the extra dame the White stones cannot connect because of connect and die. In that case Black can fill that dame anyway, and if White connects all the White stones will have only two dame and Black can kill.
Comme ca. But this cannot be seki, either.
2) The other White stones cannot connect to these White stones, and White shares the extra dame with other Black stones. And killing the disconnected Black stones means that White cannot fill both of the shared dame with the other Black stones. That is, those Black stones gain a dame. That allows Black to capture or kill the other White stones. But since there are two shared dame, Black can do that, anyway (possibly by attempting to connect and taking away a White dame), at the cost of allowing White to kill the disconnected Black stones.
If Black can make the kill without sacrificing another stone, there is a seki only when the mutual kill produces a local score of zero. IOW, the seki is unstable.
Can he do so? Say that he tries, and White fills one or more shared liberties to prevent it, finally reducing them to only one. Now Black can play in the bamboo joint, reducing White's liberties to two and killing the White stones unless White can put Black into atari from the other side. But we know that that is impossible, because Black could effect the kill by playing in the bamboo joint to start with. All we have done is to transpose moves. So Black can do so, and any seki is unstable.
(Later): No, there's a flaw here. White now has the option of killing the other Black stones while allowing the connection. White may not be able to do that if Black starts with threatening the connection. :-( Still, we're close to a proof or a construction, I think.
(Still later): Hmmm. In this case, where Black's attempt to kill the other White stones fails without reducing White's dame by attempting the connection first, can White not attack Black from the other side, forcing Black to make that play, anyway? The result could be an exchange, dissolving the seki, or life for both, leaving a seki in between. In either case, the seki is unstable, no?
wms: I'm not sure I understand all of your reasoning, Bill...a lot of the time you talk about "the black group," etc., when there are several black groups in the diagram, and I'm not sure which one you mean. The reason why I only care about Japanese in this case, and not other territory scoring rules, is because I'm really trying to determine whether or not KGS/CGoban scores games correctly when all nonnegative points are filled. Since Japanese is the only territory scoring system that KGS/CGoban supports, it is the one that is of the most interest to me :-). But your position was very interesting, because it does have the case where white killing one pair of black chains lets the other pair of black chains go from seki to a living group.
What if both of the pairs of black chains each had another white group behind them, where if the pairs of black chains became completely alive, they would be able to kill the "back" white groups? Then white will get 22 points by killing two of the black chains, but then black gets two points (for making life) and x points (for killing a white group that was in seki before the black stones made life). Would that be possible? In that case we may indeed have a position where for white to move costs white points (by letting black make life, and kill a large white group), but for black to move costs black points (by letting white kill all the black seki stones).
Mef: wms - Is something like this what you were thinking of? This one doesn't quite work, since B connecting just makes another seki, but I'm sure Bill can come up with something....
Bill: I think I can streamline the argument for the case where the question is the life of other White stones ( and friends)that cannot connect to the main White group vs. full life for the Black stones in the corner. That involves a possible semeai between the
group and the intervening Black group (
and friends).
In that semeai Black has 3 external liberties, because White has to play (or the adjacent point) before playing
. (If White plays
right away
kills.
There are 2 shared liberties between the main White group and the group. If there were N shared liberties, the
group would have N+1 liberties for the semeai.
One condition for seki in such cases is that enables Black to win the semeai. Actually, it reduces Black's liberties for that semeai by 1. So these cases cannot be seki.
To recap the argument, there are three cases.
1) Where White's capturing some Black stones (by playing on one of two potential connecting points) allows others to make full life. Example: Position 3. Such cases may be seki-like, in that the position can remain on the board with correct play until all the dame are filled, but the local score is in White's favor, and, by Japanese rules, White must play them out to get that score.
2) Where White's capturing some Black stones as above allows some White stones to be captured or killed because they can no longer safely connect to the main White group. Example: Position 4. These are not stable seki, because
a) Black can threaten that connection earlier (as in the example), or
b) if Black cannot afford to threaten that connection because it reduces one of Black's libeties, White can connect and take away that liberty earlier.
3) Where White's capturing some Black stones as above allows Black to capture some White stones that are separated from White's main group. Example: Position 5. But that capture cannot aid Black's capture of the other White stones because it takes away one of Black's liberties of a Black group adjacent to White's main group.
Conceivably there are more complex positions combining these elements, but they will either be unstable, or if stable, positions that White should resolve at the end of the game to get a favorable local score.
1) Where White's capturing some Black stones (by playing on one of two potential connecting points) allows others to make full life. Example: Position 3. Such cases may be seki-like, in that the position can remain on the board with correct play until all the dame are filled, but the local score is in White's favor, and, by Japanese rules, White must play them out to get that score.
wms: Can you explain why the local score is always in white's favor in such cases? This is what I keep getting stuck on. Let's say there are some small black groups, and some big ones. We've all seen cases where black can choose which to save - the small or the large. Can there be a situation where, when white tries to capture one of the groups, that reduces white's liberties in a way that gives black the choice on which to save - but if black maves the first move, trying to bring his groups out of seki, that gives white the choice on which black group dies? I can't find such a shape, but I can't understand why it is impossible either - and if such a shape existed, then the local score might not be in white's favor, it might be better for white to leave everything in seki.
Bill: Let me see if I understand the problem.
In this position White will certainly choose to kill the Black groups on the left rather than on the top.
But what if there were some position where White cannot choose to kill those groups, but only the groups on the top. And in addition, Black cannot save those groups, but only the groups on top. Then that would be a seki. Is that the problem?
If so, how does Black save the top groups? By threatening to live with the groups on the side, which White kills. But if Black cannot make them live, even with the move, surely White can kill them, given the move.
But maybe that move does not kill, but only preserves the seki. But that move has prevented the larger, left groups from making full life, possibly at the cost of a liberty. White still has at least three liberties left, and can kill the top groups, wiping out everything. If White has only three liberties left, however, Black can connect the top groups, killing White unless White kills the left groups. The White move does not preserve the seki.
Scatee?:Cgoban counts circles as points.
yoyoma: The two central dame can be safely filled by either white or black (the other player will respond by filling the other). Then Cgoban will score it correctly.
Notochord: I think that you have laid all the necessary ingredients on the table. Would this beast work out? If either side tries to take one of the sets of groups in exchange for the other, he ends up getting a group that is devalued. If black tries to take the four stones instead of connecting, white gives atari to that group, and plays up top, taking three groups out of four (or if Black plays on top, White gets an extra captured black stone on bottom). If White tries to approach by adding to the four, he is just captured, not a hair to Black's loss. I may have done poor arithmetic here, but we have the freedom to add or subtract increments of two to the score here (by adding white/black stones to the dumplings). Also, I assumed that kgs doesn't penalize for having more stones in seki than the other? If not, our degree of freedom can still be brought to bear, I think.
wms: Very, very, interesting, notochord! I worked it out, I get these options:
- Both players pass: B+158
- Black tries to connect: B+157
- White tries to kill: B+158
So you have a situation where black would prefer to let the seki stand than connect through and settle it. For white, it is basically just a huge ko threat - there is no profit or loss in trying to settle the seki. I tried tinkering with it, and I could shift who gets to treat it as a ko, but couldn't make it -1 for both to play first. I'll have to look at this and see how it matches what arno and bill have been saying - maybe (probably?) I'm being thick, but I just keep seeing assumptions in their arguments that I don't see as being clearly true. This looks close enough, if it can be tipped one more stone, then maybe it could be bad for both to move first while keeping the bamboo joint connections.
Notochord: I read a score of zero (outside aside) if both players pass, since all the stones seem like they would then be counted in seki (does the eye of a group in seki count?). The difference between 'White tries' and 'Black tries' seemed to be that in the first situation, White must play three times on his side: once to cut, once when under atari and once to prevent the group surrounding the four stones from living (to touch the group surrounding the four stones is suicide for White's big group). If Black tries, then he ends up playing a stone on white's side in place of the third white stone, and so black is two points worse off than the situation when White makes first move, leaving room for one score to be positive and the other negative.
Bill: Very nice, Notochord! :-) By adding a couple of White stones you can even get a positive score for Black inside the seki-like complex. You exposed my error.
wms again: On second look, it looks like in this case, black must connect through. If the situation stands this way at the end of the game, then all the black stones in the "seki" die. Am I right? Because White can choose which to kill (either the top two black groups or the bottom two black groups), and by japanese rules, that means that all four of the "seki" black groups are dead at the end of the game! That changes things a lot, I hadn't thought of how that affected this challenge...suddenly it becomes trickier.
Bill: Bill, I think it's OK. Whichever Black groups White chooses to kill, doing so enables new Black stones to arise on the board that are alive. So White's claim that all of the Black groups are dead does not stand.
I was wrong about the local score being in White's favor. But I think that this kind of position (a choice of Black groups to be fully alive) still does not have to be considered seki. The reason is that the score remains the same under area scoring, even though there is a one point difference by territory scoring, depending on who plays first. You could get a greater difference if there is a ko involved, but you are not contemplating that, are you? If you are, maybe we can construct something with a two stage ko.
wms: I'm pretty sure that in fact all four black "seki" groups are dead in the given situation. There have been several variations of Japanese rules of course, but in general at the end of the game white can claim the top stones are dead. To prove it, they play it out, white kills the stones, proving his point. Then the board reverts back to the position before playing it out, but the stones (although replaced) are still considered dead for scoring purposes. Then White can claim that the bottom stones are dead, and prove it in the same way. Thus all the black stones are dead. Note that black can only claim that the two four stone groups are dead; and in that case, there is a dead white group inside a dead black group, so the territory is white anyway.
And you are right, I do not want kos involved! Kos are always either resolved at the end of the game, or are some form of eternal ko, in which case the score just doesn't matter because the game has no result at all.
Notochord: I'm no ologist of rulesets, but I think that this makes the task trickier in the infinite sense. One could probably (with nontrivial difficulty, no doubt) prove from first principles that no such "can't connect through open gap" seki can exist, since the (should I say topological?) space that we can work with becomes so much smaller; no back groups within black's seki groups, really, PLUS more restrictions. I think that if this is the case with the japanese ruleset, then you can probably either eek out the cogitation (or the petition to a math major) to prove, or much easier, very confidently go without proof that your code works fine. Or say very extremely confidently that the probability of such a position arising to be miscored (without co-operative play) is on the order of your server mistakenly flipping a bit and miscoring, or worse (better?): game can't be scored on account of end of the world :-)
wms: Notochord, I made a mistake in my analysis. I forgot that white had to make the bulky 5 shape to kill a black group on the top or bottom (middle of edge) after cutting! So you are right, in that assuming this is a seki, then we have exactly what I asked for - a seki situation where we have two one-eyed groups separated by bamboo joints! It really is a very clever solution, I couldn't find anything like it. And you have demonstrated to Arno and Bill that in fact, it is possible, despite their reasoning otherwise. I am prepared to give you a 3 month KGS Plus membership in return for answering my puzzle, if you wish. :-)
"Notochord: I believe that I wish :-) Thanks, wms! (My KGS account is under the same name)"
However, it seems that the KGS scoring system is probably safe, which is also good news! There are two reasons why Notochord's amazing position doesn't break the scoring system:
- As discussed above, I'm pretty sure that in fact black must play in this position, sacrificing top or bottom to connect through, or else by Japanese (fairly convoluted) rules all 4 groups are dead.
- Even if that isn't necessary, KGS will score correctly, because the white groups inside the black eyes are alive. The KGS scoring system would consider the black groups to be fully alive if they had eyes with no living stones inside. In this case, black cannot kill the stones in the upper left and lower left corners, and thus KGS will not get confused.
I'm pretty sure now that the KGS scoring system is "safe". I think that issue 1 above keeps it that way - the only way I can imagine such a situation is where you have two of these "connections" and cutting one will let the other live. As long as it is possible for one player to choose which side lives and which dies, by Japanese rules, the non-choosing player will have to play to force a choice to be made or else lose both sides...thus one of the players should keep playing in the position until it is no longer seki.
But although I'm 99.9% sure that the score system is safe, I could be wrong. :-) I will happily give out another KGS Plus membership if somebody manages to find a position which fixes the two issues with this one...or if you manage to fix issue #2 and if you can show that I am mistaken about how Japanese rules work in a position like this.
By the way, there are several bestiaries of weird seki floating around. It's fairly abstruse, but I'm pretty sure none has a position like yours, so I think you have truly invented something new here. :-)
Bill: Congratulations, Notochord! I misread the position, too.
= pass.
I overlooked the Black pass after . Within the broken seki Black has 48 points, White 47.
Notochord: I think that at least under Japanese scoring, having capture
will force White to take a loss. Black holds the prisoners in his hand, so I think that White will be better off breaking the seki, going from a position valued at +5 because of the prisoners to one which is the same as if White began with
, except for the addition of a black and a white stone to either Black's or White's territory, which cannot alter the score: +1. I think that there might be some trickyness here, too? If White passes after
capture, then Black should pass, too, since if he plays at the center of the eyeshape to live the group, White can then settle the shape so that by tewari, Black has played a superfluous move inside her own territory: +1 -1 = 0.
Bill: If captures the 5 White stones, forcing White to play a nakade, then
passes and play continues as above, with the same net result (B +1). What Black gains in captives White gains in territory.
Notochord: Sorry if I'm not being clear. What I mean is that the capture should force White's hand in settling the seki. If Black passes, then can't White pass too and leave it as seki :+0? Or am I not thinking of the correct rules? There seems to be nothing flawed (from the is-it-seki standpoint) in passing since the onus is still on White to gain something, from moving first, but wouldn't
capture be the optimal response?
Bill: Thanks for the clarification. I thought you meant a loss by comparison with the diagram.
If Black plays first Black still gets 48 points, but White gets 49.
Neither player can afford to play first, so it is seki. :-)
wms, take a look at the first diagram. White indeed can kill the top groups of stones. However, that allows Black to play two new live stones ( and
). Therefore the Black stones on the top are alive. Similarly for the Black stones on the bottom, if White killed them.
Here is the relevant clause:
Article 7. Life and death 1. Stones are said to be "alive" if they cannot be captured by the opponent, or if capturing them would enable a new stone to be played that the opponent could not capture. Stones which are not alive are said to be "dead."
- wms: I think I've seen translations or interpretations of this that indicate that by "a new stone to be played" they mean as a replacement for one or more captured stones. So for example, in a standard 2-stones-take-1-enemy-stone-to-become-empty-triangle-in-atari snapback, the single stone is alive - if it is captured, then it can be replaced with an unkillable stone. In Notochord's example, the unkillable stones are not replacements for captured stones, so they don't help black.
Bill:
One official example is this seki.
The
stones are alive because if White captures them,
Black can play two new stones,
and
, that are alive.
I'm not sure what you mean by issue #2. Do you mean the KGS scoring system getting confused? Can it handle Notochord's example plus the next two slight variations?
wms: Whoever wrote this, I don't know what you mean when you say you don't know what I mean. :-) It is quite simple. For KGS to become confused, you need a position like notochord's, but with no living stones inside the seki eyes. The positions you give don't even seem to be at the end of the game - you talk about one player or another having to move - so they don't matter at all.
Bill: Does the KGS scorer require that the players make all necessary protective plays? Thanks.
wms: Not sure what "all necessary protective plays" means. What is a protective play? To guarantee correct score, KGS requires you to fill all dame that can be filled without costing you points. 99%+ of the time, you don't even have to do that, but if you try to score a game on KGS and see that (after correctly marking which stones are dead) it misjudges false eyes, then you press undo, fill the dame, then score again and it will be correct.
Bill: So if those dame are filled, the only stones remaining with dame are dead or in seki. And the dead stones should be marked, right?
wms: Yes, but interestingly enough, you cannot always spot seki groups by dame! There are cases where a chain in seki will have no dame directly bordering it. Instead it will have a false eye separating it from the dame.
Bill: Right. But then can't all of these examples be recognized as being in seki because of the dame?
wms: Perhaps, but that isn't the algorithm I use. I count eyes instead. The problem with saying "dame means seki" is that then you must fill all dame to have any hope of a correct score, which would irritate a lot of players who would prefer to spend as little time as possible filling dame. So I count eyes instead, which means you only have to fill dame points if they are in a bamboo join or other such position.
(Sebastian:) This Challenge was originally posted by wms in the KGS Go Cafe. Please feel encouraged to find a different name for this page. I'm not even sure if the "s" in "wms's" is correct.
- wms: "wms's" is correct if by "wms" you mean my full name. If you mean just my initials by "wms" then "wms'" is correct. I don't think it's a big deal.
- (Sebastian:) That raises the question: What characterizes you better? ;-)
- wms: It doesn't matter. I answer to both. "Bill" is fine, and is what people actually call me, but "wms" is more explicit I guess since there are quite a few othel Bills who play go.
Arno: I don't know if the following is a proof that such a position cannot exist, but maybe it is a hint for other people:
As far as I can tell these are the only possible postions, as the requirement says: two dame points which have the potential to connect the two black groups.
- let' call the concerned groups
,
(the one eyed seki groups) and
,
(the surrounding white groups on either side of the dame points); for here onwards dame refers to the critical two dame points in question.
- by definition each group has at least two liberties:
-
,
have at least two liberties: one liberty which is dame, and one eye
-
,
have at least two liberties: one which is dame and at least one other liberty (otherwise they are in atari and Black just captures).
-
- If Black moves from
or
- the liberty count for
/
is not reduced (the filled dame has at least one more liberty adjacent to the other B group)
- White loses a liberty of at least one of
or
.
- the liberty count for
- White has to move on the other dame in order to prevent Black from linking up and making life:
- Black loses a liberty on each group
,
.
- Either White's play connects
with
or the W group (the one from which the move extends) loses a liberty.
- Black loses a liberty on each group
- As we look for seki the position must now satisfy: liberties(
or
) > liberties(
or
) (otherwise Black can capture one of the two groups
,
and thereby connect both of his groups, as
/
are cutting groups in regard to
/
)
- or to be more exact in the initial position (not counting the dame points!):
- bamboo joint: liberties(
+
) > MIN(liberties(
), liberties(
)) (
+
denotes the connected group)
- diagonal joint: MIN(liberties(
), liberties(
)) > MIN(liberties(
), liberties(
))
- bamboo joint: liberties(
- With the result from above, let us assume that White moves from
or
first:
- as White has more liberties she can just fill one of the dame points and kill one of the Black groups (the one with less liberties) and therfor the other one as well.
So is this a proof? I don't know. Maybe I have overlooked something?
wms: You assume that white having fewer liberties than black means that white can always be killed. I'm not sure that is the case. If white has outside liberties which are part of another seki position, then it is possible that black cannot fill all those outside liberties, in which case and/or
might actually have fewer liberties than
and/or
, but it will still be impossible for Black to capture. Am I right? Or is there some reason why this is not possible? Or is it possible but irrelevant for some reason?
Arno: I would argue that it is not as you mention: if the white groups are caught in another seki, then the liberties of Black and White groups there must be equal[1]. If you play the dame, the liberty count for White reduces (by 2 in case of bamboo joint, by 1 each in case of the diagonal joint), so the other seki should get destroyed as well and Black can start filling liberties, no? I know my train of thought sounds to easy to be true. I also have doubts that it is that easy.
On a side note: unless /
are caught in another seki, White can always play as to reduce
/
to two liberties. This is because, by definition (no seki) White is able to surround
/
completely. This raises the bar for the sought after freak position (some sekis interwined): as
/
cannot be caught in another seki with less than two liberties this means that
/
have to be in another seki as well. Sounds ever more impossible.
[1]: not really equal as that seki might not have been played out. But in that case, Black should play out the seki first (i.e. play until there are no more moves that would keep the position a seki) before starting on the dame points.
aib: If black has two one-eyed groups, then connecthing them will result in unconditional life. For black to get in trouble when attempting to connect, white has to play (inside the dame). However, it has already been proposed that playing inside the dame damages white (else she would have already connected). Thus white cannot prevent the black groups from connecting and getting life.
This would explain why the closest we've been able to get is with two dames (diagonal empty spaces as in Scatee?'s example).
As long as black has a clear path between the two one-eyed groups (i.e. a dame), a black move will not change the situation (i.e. will always result in at least one liberty). Thus, black will always be able to connect.
Does that seem logical? It does to me.
Imagist: The situation in the top left is a seki in which neither player benefits from playing a. If black plays a, black loses the game by 1.5 points. If white plays a. black starts a round robin ko which ends in no result: a failure for white. Therefore neither player plays a and the top left is seki.