I took the screenshot a couple of months back, it was most probably from tsumego hero
I like to think I’m reasonably solid on capturing race theory, but I have one area of knowledge fuzz—this puzzle is an example. It’s obviously eye vs no eye, so the shared liberties go to Black. E9 is what makes me uncertain. It’s a shared (perhaps internal is the correct term) liberty, but is also half a kosumi for White.
How are the liberties at E9 / F8 counted? Is this even knowable?
You need to ask yoursef two questions:
Edit: sorry I was confused. In this situation, it’s eye vs no eye, so no seki is possible. Either Black dies or White dies. And if White wants to kill, White needs to play E9 at some point anyway. And if Black wants to kill, Black needs to play F8 anyway. So you can imagine that the exchange F8-E9 has already occurred and count liberties from there.
Here is something similar, Black to play.
Probably in that situation Black should tenuki but suppose Black plays E5. If White connects with F4 then White dies. So after Black E5, White can tenuki and it’s seki. Black may capture three stones later.
Black F8 - White E9 is a privilege or White E9 - Black F9 is essentially a reverse privilege so we may as imagine it to be exchanged.
Black A14 is an option to create a ko so as follows let us first analyse assuming that Black does not use it.
Black has 2 eye liberties, 4 external approach liberties (including A14, White A13 first does not reduce the excess approach liberties, White to start starts at A14) and the 2 internal liberties E10 + E11, altogether 8 approach liberties.
White has 7 direct external liberties (excluding the ignored F8) and the 1 extra external liberty A9, which Black can fill last or else needs to approach via the 1 additional approach liberty A7. If Black simply can fill it last, White has altogether 8 approach liberties. If, however, Black needs to approach via the 1 additional approach liberty A7, White White has altogether 9 approach liberties.
So we need additional theory for the type of capturing races “eye versus no eye, the no eye group has exactly one external approach defect” (here A9 possibly to be approached at A7). (If there are at least two approach defects, Black would need to approach at least every additional approach defect.)
Suppose White to move. Suppose White plays the 4 external approach liberties and then the 2 relevant internal approach liberties so makes his first 6 approach moves. Meanwhile Black fills 6 of the direct external liberties. The remaining liberties are 2 black eye liberties versus 1 white direct external liberty and 1 white external liberty of the approach defect. White moves next and wins the capturing race. Black is too late for possibly filling A9 last. Since White starts at A14, Black cannot choose the ko option.
Suppose Black to move. Suppose Black does not choose to create the ko. Suppose Black fills the 7 direct external liberties. Meanwhile White plays the 4 external approach liberties, then the 2 relevant internal approach liberties and then 1 of the two eye liberties. This leaves 1 eye liberty for Black versus the 1 white external liberty of the approach defect, which Black to move next now fills last winning the capturing race. Therefore, Black need not choose the ko option to win the capturing race and early during the middle game choosing it nevertheless for better endgame is inconceivable.
Hence, besides additional remarks on the ko option, theory for this type of capturing race is: If there is an equal number of approach liberties without counting any extra approach move in the one approach defect of the eyeless group, the eyeless player to start wins while the eye player to start wins by playing the approach defect’s physical liberty last.
This is the key for me, I was zeroing in on this idea and started typing something along those lines while drafting the OP…it felt icky though. Once I imagine the F8-E9 exchange played, the rest is a doddle. Cheers.
