An unremovable ko is a ko that one player can start at any time, while the other player can do nothing to remove that option.
In this position, white has an unremovable throw-in ko. White can, at any time, play b to start a ko for the corner. Black can do nothing to remove this ko.
It is not an unremovable ko the other way around. Although black can start the ko at any time by playing a, white can remove the ko by playing a himself, creating seki.
Black cannot remove this ko by filling, as he will die. On the other hand, White can take and fill the ko for seki.
These two positions are well known. See Sending three, returning one.
When neither player is komaster, usually the player with fewer stones at stake (White in these examples), should start the ko. Even if he loses the ko, he will get compensation elsewhere in the ko exchange. The same is usually the case when Black is komaster. (The fact that Black cannot remove the ko is no disadvantage to Black then.)
When White is komaster, however, Black is at a serious disadvantage, because he has no effective play to remove the ko. The longer he waits to start the ko, the less he gets in compensation for losing it. Therefore, as paradoxical as it may seem, Black should usually start the ko as soon as possible.
If an unremovable ko remains at the end of the game, it may have implications for the status of groups (depending on the rule set):
Black claims: They are dead.
White says: Show me.
Play goes as shown. After , white can live with a move at , or retake the ko with a move at .
Since the ko is bigger in this position, black takes with at a and white lives with at .
So what is the status of the white stones in the upper right?
Bill: The White stones are dead by Japanese rules, and also by Korean rules, I believe. By area scoring, optimal play is as given. By Lasker-Maas or Spight rules (territory rules), the given play is properly done in the encore.
Erik?: Japanese rules state: '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.' The latter seems to be relevant here.
Bill: If Japanese hypothetical play goes as above, Black kills the White stones with and White cannot take the ko back, because White has not passed for it.
Herman: Thank you, Bill. Which leads me to the following question:
In this situation, is optimal play by black to claim seki?
Bill: I think that (under area rules or certain territory rules) Black scores 1 pt. better by "killing" the right side group in exchange for the ko.
Play could stop at this point by agreement that the White stones on the right are dead, and the score would be 25 - 24 in favor of Black. Or White could make Black capture those stones, for the same net score. If Black does not play in the encore the result is jigo.
Erik: I discovered the position when I was analyzing solutions for 4x6 Go. Here's the original position:
BTW I'm curious if this was already a know rules-beast, or if I was actually the first to stumble upon it.
Herman: I had never seen it before, and I've seen a lot of rules beast already, so I wouldn't be surprised if this is a new discovery.
Bill: It's new to me. :)
Herman: I found a similar position at: http://www.goban.demon.co.uk/go/bestiary/molasses_ko.html (third diagram)
Herman: I tried to build a position like this (because you told me you'd found it on 4x6), but failed. In the given positions, after and , black would capture with at a. So white should play at a, resulting in the following:
Herman So now is the ko threat white needs.
asmobia?:Usually an unremovable ko is only for one side. However, a person called 序列號 (which is a username instead of a real name) discovered an unremovable ko for both sides in 2008.
Area counting, komi is 0.5 point.
Black has 38 points and White has 37.5 points. The player who wins the KO (at bottom left corner) wins the game.
Black can never fill "A" after winning the KO, otherwise he dies.
On the other hand, White cannot remove the KO either. If White wins and fills the KO, then Black doesn't need to keep any KO threat and Black can capture two white stones at above. After all, Black gains more (4 points) than White does (1 point).
This beast challenges Ing's and Spright's rule, those rules assume every "single KO" can be removed if playing reasonably, which is not true.
? I don't get it. Can't Black just take the two white stones, let White have the ko, and win the game?