One Thousand and One Life-and-Death Problems / Errata

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

This table serves as a table of contents: click the problem numbers to read our contributors’ observations
Annotations to problems
Problem
number
Page Type Error /
Remark
2nd printing
(2006)
Discussion
page?
Live in one 1–200
12 4 answer diagram B misplaced same
75 20 Yes
87 22 Yes
92 24 Yes
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
451 116 alternate continuation another B3 same
572 144 status seki possible (not “two eyes”) same
579 146 Yes
581 146 Yes
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
719 178 Yes
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[1] 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
871 212 alternative solution same
872 212 alternative inferior solution leaves a bigger threat to live same
881 214 alternative inferior solution same
903 218 alternative solution same
910 220 alternative inferior solution same
929 224 alternative inferior solution same
937 226 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

[1] The introduction says no correct solution involves ko, apart from double ko.

Errata

Page 4, problem 12, solution diagram

[12]

[Diagram]
Problem 12 solution  

The solution is correct, but the location of the 'b' annotation is wrong, it should be one line lower at the 1-3 point.



Page 28, problem 108, solution text

[108]

The solution text says: If White plays the diagonal move..., but it should be: If Black plays the diagonal move....



Page 42, problem 158, solution text

[158]

The solution says: ...a dead four-point eye space., but it should be: ...a dead five-point eye space.



Page 44, problem 168, solution diagram

[168]

[Diagram]
Solution diagram  

"If Black turns at 1, White can't stop him from getting two eyes."

[Diagram]
White can make seki (6 at 4)  

However, White can obtain a seki by playing with W2 and W4.



Page 44, problem 172, solution diagram

[172]

[Diagram]
Solution diagram  

"Black connects at 1 and makes an eye. There is nothing White can do."

[Diagram]
White can make seki  

White can make a seki by playing W2, making 'a' and 'b' miai.



Page 50, problem 195, solution diagram

[195]

[Diagram]
Book Solution  

The book claims that B1 makes a seki.

[Diagram]
Error  

However, W2 and W4 turns the position into a ko.

[Diagram]
Correct Solution  

Therefore, the correct solution should be to connect at B1 instead. It guarantees at least a seki.

[Diagram]
Seki  

W4 and W6 can turn the position into seki in gote.

[Diagram]
Independent life, plus  

W6 takes back at white+circle.

Or Black can make independent life, also in gote. Later B3 - W6 is sente.

Page 63, problem 243, problem diagram

[243]

Corrected as below in second printing (2006), where this is problem 245

As presented, black can't kill white. The solution presented can be refuted with white 2:

[Diagram]
 



Likely, the intended problem needs one more black stone to the right:

[Diagram]
 



Page 77, problem 300, problem diagram

[300]

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.

Page 82, problem 316, solution diagram

[316]

[Diagram]
Problem 316  

"Black peeps at B1. 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 circle.



Page 86, problem 327, alternative continuation

[327]

[Diagram]
Problem 327  

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

  • after B, Black can answer by removing White’s stones;
  • after C, Black needs an extra move to remove them.

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.



Page 86, problem 329, solution text

[329]

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.

Page 96, problem 372, solution text

[372]

[Diagram]
Problem 372  

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.



Page 98, problem 376, solution text

[376]

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.

Page 100, problem 386, solution diagram

[386]

[Diagram]
Solution  

The text says "Black B1 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.



Page 102, problem 389, solution text

[389]

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.

Page 106, problem 408, solution text

[408]

"corner" rather than "cornert". (This has been corrected in the 2006 edition.)

Page 112, problem 430, solution text

[430]

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.

Page 116, problem 451, solution text

[451]

[Diagram]
Problem 451  

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."

Page 144, problem 572, solution diagram

[572]

[Diagram]
Solution  

"With B1, Black gets two eyes because it makes miai of the points W2 and B3."

[Diagram]
Seki  

Obviously white can just play W4 to make a seki.



Page 147, problem 593, solution

[593]

The book's solution shows a variation where black lives:

[Diagram]
Book solution (life for black)  


But if white plays 2 at 3, the result is a seki:

[Diagram]
Seki variation  



Page 154, problem 621, solution diagram

[621]

[Diagram]
Problem 621 -- solution  

The solution given is B1, but the comment says that B1 at a also works.

[Diagram]
Problem 621 -- alternate  

The book is wrong. a and b are miai for White to live.



Page 181, problem 738

[738]

[Diagram]
Black to Live  

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.



Page 182, problem 740, solution text

[740]

[Diagram]
Black to Live  

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.



Page 183, problem 751, problem diagram

[751]

[Diagram]
Problem diagram  

"Black to live"

[Diagram]
Solution diagram  

"Black plays B1, sacrificing two stones. After White captures with W4, Black B5 makes a second eye."

[Diagram]
The error  

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 white+circle stone should be removed from this problem for the intended solution to be the best solution.



Page 184, problem 752, solution diagram

[752]

[Diagram]
1  

"B1 is the vital point. Black B3 and B5 are obvious. Black has two ways to make his second eye."

[Diagram]
2  

Instead of W4, White can extend at B5 in Diagram 1, and Black does not have enough liberties to play at W4, and so must capture the four white stones...

[Diagram]
3  

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



Page 192, problem 787

Wrong: White resists with 2 and 4, but Black ends up short of liberties.
Right: White resists with 2 and 4, but ends up short of liberties.

Page 192, problem 789, solution

[789]

I believe that the result is seki, not two eyes for black.

[Diagram]
Seki  



Response: In the book solution, Black has 1.5 points and White ends in sente. In the seki variation above, Black has 0 points and White ends in gote. Unless it is the very end of the game, it is better for White to take sente than to gain 1.5 points in gote.

Page 194, problem 799, solution

[799]

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:

[Diagram]
Claims that black can win with no ko  

But if white plays W4 at B5, white can make a ko! Watch:

[Diagram]
White can force a ko  

And now we have an approach ko for white, which, if won, kills black.

(B5 at W6 is a direct ko.)



Page 212, Problem 871, problem diagram

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.)

[Diagram]
 



Page 231, problem 966, problem diagram

[966]

[Diagram]
White lives  

The problem says "Black to kill", but White can live with this sequence.



Multiple solutions

Page 36, problem 134, solution diagram

[134]

The solution given is:

[Diagram]
Problem 134 solution  

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:

[Diagram]
Problem 134 solution  

Black easily lives with 1 or B. Black A also lives, but with a double ko.



Page 64, problem 244, solution diagram

[244]

[Diagram]
Solutions  

Solution diagram gives B1 as the solution. However, Black a works too.

PJT I think B1 is usually the optimal solution, as it leaves only one threat to live: W2 at a (or b, but a very faintly threatens a cut cross). 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.

[Diagram]
More threats  

Playing B1 at a leaves White four threats (W2, W4, W6, W8), and the aji of W6 and W8.

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 W2, she can follow up with W4 at:

  • B5, living with 2 + 1 = +3 points (threat value 19) and a ⅓-point ko in which she leads;
  • W4, living with 2 - 4 = -2 points (threat value 14) but keeping the cut at W6 alive;
  • B3, living with 2 points, leaving no no counter threats but several points and the cut at B6 undecided.



Page 86, problem 325, solution diagram

[325]

[Diagram]
Solutions  

Solution diagram gives B1 as the solution. However, Black a also works, and yields exactly equivalent options in terms of threats, liberties and score.



Page 88, problem 337, solution diagram

[337]

[Diagram]
Solutions  

Solution diagram gives B1 as the solution. Black a also kills. However, the book solution B1 is better than a, because if black needs to actually remove the white stones, B1 avoids any possibility of ko in the corner.



Page 198, problem 810, solution diagram

[810]

[Diagram]
Solution diagram  

5 at a also works.

PJTI believe that a is inferior because it leaves a larger threat.



Page 200, problem 821, solution diagram

[821]

[Diagram]
Solution diagram  

The solution from the book.

[Diagram]
Alternative solution  

B1 forces W2 and after B3 the points a and b are miai.

Note: This solution works even with outside liberties.



Page 204, problem 839, solution diagram

[839]

[Diagram]
Solution diagram  

5 at a or b also works, but these are slightly inferior to the book solution since they do not kill White as quickly.



Page 212, problem 872, solution diagram

[872]

[Diagram]
Solution diagram  

5 at a also works, but leaves White a bigger threat to live.

  • After B5, White b threatens to live with 6 points (including B1), leaving Black 1 threat to kill and White no further moves.
  • After a, White b threatens to live with 8 points, leaving Black 1 threat to kill and a 1-point gote at B5 (¿or should it be counted as a half-point ko?).



Page 214, problem 881, solution diagram

[881]

[Diagram]
Solution diagram  

3 at 5 also works, but is slightly inferior to the book solution.



Page 218, problem 903, solution diagram

[903]

[Diagram]
Solution diagram  

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.

[Diagram]
continuation  

Now a and b are miai.



Page 220, problem 910, solution diagram

[910]

[Diagram]
Solution diagram  

5 at a also works, but is slightly inferior to the book solution.



Page 224, problem 929, solution diagram

[929]

[Diagram]
Solution diagram  

5 at a also works, but is slightly inferior to the book solution.



Page 226, problem 937, solution diagram

[937]

[Diagram]
Solution diagram  
[Diagram]
Inferior solution  

This sequence also kills, but White is left with many internal liberties.



Page 226, problem 938, solution diagram

[938]

[Diagram]
Solution diagram  

3 at 4 also works.



Page 228, problem 945, solution diagram

[945]

[Diagram]
Solution  

The solution in the book shows this sequence as a solution. However, having B1 to directly hit the vital point at B3 is also a solution (think incomplete bulky five).



Page 230, problem 951, solution diagram

[951]

[Diagram]
Solution  

The solution in the book shows this sequence as a solution. Playing directly at 4 works too.

[Diagram]
Solution  

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.



Page 238, problem 986, solution text

[986]

[Diagram]
Solution diagram  

The solution text says White is short of liberties, so he can't approach. However, even if white could approach, it wouldn't make any difference, white would still be dead.


One Thousand and One Life-and-Death Problems / Errata last edited by DanSchmidt on February 1, 2023 - 17:51
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