Dieter: The first line to investigate is what happens when Black removes all eye potential in the connect-and die at . brings the capturing race to 3-2 in his favour. Black chases the three cutting stones and with reduces both strings to 2 liberties, but solves it all.
So something has to be done with the order of play, it seems.
This way doesn't change anything: Black captures all but the big string, for a is still connect and die.
White can capture the top string, but Black has time to capture the left string and White's eyespace is not big enough: all stones die.
So White can make life for the left and centre string but loses the rest of the stones. Incidentally, is necessary. If at , then ata.
In summary, all stones have a settled status. The marked white strings are alive, the unmarked are dead. All Black stones live but the two marked stones.
Could it be that and are a blind spot to the problem composer? Still thinking --Dieter
Herman: at a now, I guess?
unkx80: Dieter and Herman, you are both right.
Herman: Meaning, I assume, that you did not consider Dieter's and when composing the problem (but, perhaps, only the next diagram below), but that my suggested saves the day and makes this variation work as well? :-)
unkx80: Sorta yes, but I already corrected this deficiency in the final problem composition presented here. The problem composition that had this deficiency is in the /Comments page. Otherwise, I don't need to give the White chain on the left of such a long tail.
Dieter: It is I who had a blind spot. As soon as Herman suggested , I saw the two variants of the solution. Then when unkx80 said we were both right, I assumed that he was hinting at his original composition.
It is a very nice problem, really astute.
unkx80: Thanks! =)
Herman: If like this, then after this sequence the move at a and the throw-in at are miai for Black.