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.
|Page||Type||Error / |
|2nd printing |
|Live in one 1–200|
|12||4||answer diagram||B misplaced||same|
|108||28||textual||“White” for “Black”||same|
|134||36||extra (equivalent) solution||same|
|158||42||textual||“four-” for “five-” point||same|
|168||44||seki||White can make seki||same|
|172||44||seki||White can make seki||same|
|195||50||solution||Given move: ko; different move seki||same|
|Kill in one 201-400|
|243?||63||status||White can live (extra stone needed)||corrected (is # 245)|
|244||64||extra ¿inferior? solution||leaves more slightly smaller threats||same|
|300||77||text||“live” for “kill”, in section kill, easy||text corrected|
|316||82||answer diagram||C missing||same|
|325||86||extra (equivalent) solution||same||Yes|
|327||86||extra inferior solution||very marginally inferior!||same|
|329||86||textual||“White” for “Black”||same|
|337||88||extra inferior solution||same|
|372||96||explanation||“White has only one eye”, but all are false||same|
|376||98||textual||“Black” for “White”||same|
|386||100||textual||“C”, “D” for “B”, “A”||corrected|
|389||102||textual||“Black” for “White”||same|
|Live in three 401–600|
|408||106||textual||“cornert” for “corner”||corrected|
|430||112||explanation||dead, though not “short of liberties”||same|
|572||144||status||seki possible (not “two eyes”)||same|
|593||147||status||seki possible, diagram suggests territory||same||Yes (Bill)|
|Kill in three_ 601–700|
|621||154||solution||invalid alternate solution given||same|
|Live in five 701–800|
|738||181||text||“kill” problem in “live” section||text corrected|
|740||182||incorrect refutation||wrong solution dies without ko||same|
|751||183||problem diagram||stone should be removed to make sense of explanation||same|
|752||184||status||White can make a Mannen ko||same||Yes|
|787||192||text||“Black” for “White”||same|
|789||192||status||“I believe” seki possible||same|
|799||194||status||White can make ko||same|
|Kill in five 801–1,001|
|810||198||alternative inferior solution||leaves larger threats||same|
|821||200||alternative ¿equivalent? solution||works with outside liberties||same|
|839||204||alternative inferior solutions||same|
|872||212||alternative inferior solution||leaves a bigger threat to live||same|
|881||214||alternative inferior solution||same|
|910||220||alternative inferior solution||same|
|929||224||alternative inferior solution||same|
|938||226||alternative ¿inferior/equivalent? solution||same|
|945||228||alternative ¿inferior/equivalent? solution||same|
|951||230||¿alternative inferior solution?||someone else claims ko||same|
|966||231||status||“White can live”||same|
|986||238||explanation||‘“short of liberties” irrelevant’||same|
 The introduction says no correct solution involves ko, apart from double ko.
The solution is correct, but the location of the 'b' annotation is wrong, it should be one line lower at the 1-3 point.
The solution text says: If White plays the diagonal move..., but it should be: If Black plays the diagonal move....
The solution says: ...a dead four-point eye space., but it should be: ...a dead five-point eye space.
However, White can obtain a seki by playing with and .
takes back at .
Or Black can make independent life, also in gote. Later - is sente.
As presented, black can't kill white. The solution presented can be refuted with white 2:
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. Note, the second printing (from June 2006) has this correctly labelled as "Black to live".
Moreover, the problem is far easier than those preceding it.
"Black peeps at . If White a, Black b. If White b, Black c also kills White."
The c is omitted in the diagram.
Note: Black c can be both .
The solution says "Black connects with 1. If white blocks at A, Black makes a placement at B."
A placement at C also works. It seems, however, very marginally inferior: in both cases White has one more threat to live, but
If Black were weaker on the outside this could be significant, but in the given (probably unrealistic) situation it is almost inconceivable that it could matter.
The solution says If White hanes at 1, White can't get two eyes. It should read: If Black hanes at 1, White can't get two eyes.
The solution says "Black 1 links up with his three stones below. White has only one eye."
Actually, white has no eyes; both the 1-2 point and the 2-3 point are false eyes.
The solution says After Black throws in at 1, there is nothing Black can do. It should read: After Black throws in at 1, there is nothing White can do.
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."
This has been corrected in the 2006 edition.
The solution says Black 1 exploits White's shortage of liberties. Black can't make two eyes. It should read: Black 1 exploits White's shortage of liberties. White can't make two eyes.
Solution text says "If Black 1 at 2, White will descend to 1 and the black stones are short of liberties." Actually, the black stones are not short of liberties, but Black is dead anyway.
Solution says "...If White 2, Black draws back to 3 and he is alive."
Should say "...If White 2, Black at 3 or a lives."
Obviously white can just play to make a seki.
The book's solution shows a variation where black lives:
From the first printing dated February 2002.
The solution diagram shows how Black kills White. This problem is misplaced and belongs in part six of the book which contains five move Black to Kill problems.
The given solution is correct. However, the solution text then says If Black 1 at 3, White 1 leads to a ko. This is wrong, there is no ko; if Black 1 at 3, White 1 kills.
Obviously, the stone should be removed from this problem for the intended solution to be the best solution.
" is the vital point. Black and are obvious. Black has two ways to make his second eye."
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...
I believe that the result is seki, not two eyes for black.
Black is not unconditionally alive. (Page iv says that unconditional life is intended in all "Black to live" problems.)
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 marked black stone is missing in the original diagram; without it there is no way to kill white. (This has been corrected in the 2006 edition.)
The problem says "Black to kill", but White can live with this sequence.
The solution given is:
Black easily lives with 1. Black A also lives, but with a double ko.
However, there is another solution equally as good as 1. The solution should look like this:
Black easily lives with 1 or B. Black A also lives, but with a double ko.
Solution diagram gives as the solution. However, Black a works too.
PJT I think is usually the optimal solution, as it leaves only one threat to live: at a (or b, but a very faintly threatens a cut ). This threatens to take 6 points of territory by capturing at b (or a); since killing White makes 16 points, the threat is worth 22 points.
Playing at a leaves White four threats (, , , ), and the aji of and .
However, it is possible that in some ko situations a is better, as here White’s first threat seems smaller. If Black does not answer , she can follow up with at:
Solution diagram gives as the solution. However, Black a also works, and yields exactly equivalent options in terms of threats, liberties and score.
Solution diagram gives as the solution. Black a also kills. However, the book solution is better than a, because if black needs to actually remove the white stones, avoids any possibility of ko in the corner.
PJTI believe that a is inferior because it leaves a larger threat.
forces and after the points a and b are miai.
Note: This solution works even with outside liberties.
5 at a or b also works, but these are slightly inferior to the book solution since they do not kill White as quickly.
5 at a also works, but leaves White a bigger threat to live.
3 at 5 also works, but is slightly inferior to the book solution.
5 at a also works.
Comment: If 5 at a, 6 at b lives. But 5 at b seems to work.
Comment on Comment: 6 at b doesn't work, see next diagram.
5 at a also works, but is slightly inferior to the book solution.
5 at a also works, but is slightly inferior to the book solution.
The solution in the book shows this sequence as a solution. Playing directly at 4 works too.
Actually, whoever wrote this is incorrect. If black plays at four first instead of one, black must draw back to place two white stones in atari, otherwise white lives outright. This reduces black's liberties by one and allows white to throw in for a ko due to black's shortage of liberties. It's a difficult ko for white to win, but it's still a ko nonetheless.