It is of course completely possible that the readers are mistaken about the 'mistakes', so please add if you know more!
Thanks to Bozulich for this great book. The lack of explanations in the book really motivates one to look further and work on the problems in depth.
--Skelley
Table of contents |
Warfreak2: As far as I can see, this
kills. B cannot make his two eyes, as when he captures at a, the marked point is and under-the-stones atari for W, killing black.
Warfreak2: Actually I tell a lie, B ignores the atari and plays to make an eye, then recapturing W in snapback.
KarlKnechtel: No, you were on to something. W should instead play to the right of the marked point. B then cannot play the marked point due to damezumari, and thus dies being unable to divide the space.
RoryDix?: In conclusion, the solution given in the book is perfectly correct and is the only move for black to live.
Problem 300 is described as "Black to kill" and is in the section labelled "One-move problems - Black to kill" when it is actually Black to live.
"Black peeps at . If White a, Black b. If White b, Black c also kills White."
The c is omitted in the diagram.
Note: The c can be either of the two points.
The text says "Black creates a dead eye space. If White a, Black b. If White c, Black d."
However, there are no c and d on the diagram.
It is possible that the intended sentence is "If White b, Black a."
But canīt Black get an Seki ?
mdm: A seki still means White is alive. But I think the book solution is a better result for Black - more points and sente.
Actually, Black can just connect his stones at a or b since White cannot make a false eye on the left. This way he does not have to sacrifice two stones to live.
Obviously, the stone should be removed from this problem for the intended solution to be the best solution.
Instead of , White can extend at
in Diagram 1, and Black does not have enough liberties to play at
, and so must capture the four white stones...
Now White can play back under the stones to form a 10,000 year ko. The book states none of the "Black to live" solutions involve ko...
/Problem 752 - Mannenko Discussion
I believe, that the result is seki, not two eyes for black.
The book explicitly states that there are no problems that need a ko to solve, but in this one the answer doesn't mention a move by white that forces a ko!
The solution given in the book:
And now we have an approach ko for white, which, if won, kills black.
( at
is a direct ko.)
The problem says "Black to kill", but White can live with this sequence.
Solution diagram gives as the solution. However, Black a works too.
The solution in the book shows this sequence as a solution. However, having to directly hit the vital point at
is also a solution (think incomplete bulky five).