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

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]

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

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

Page 44, problem 172, solution diagram

[172]

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

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

Page 50, problem 195, solution diagram

[195]

The book claims that makes a seki.

However, and turns the position into a ko.

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

and can turn the position into seki in gote.

takes back at .

Or Black can make independent life, also in gote. Later - 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:

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

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]

"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 .

Page 86, problem 327, alternative continuation

[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]

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]

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.

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]

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]

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

Obviously white can just play to make a seki.

Page 147, problem 593, solution

[593]

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

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

Page 154, problem 621, solution diagram

[621]

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

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

Page 181, problem 738

[738]

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]

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]

"Black to live"

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

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.

Page 184, problem 752, solution diagram

[752]

" 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...

/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.

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:

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

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

( at 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.)

Page 231, problem 966, problem diagram

[966]

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:

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.

Page 64, problem 244, solution diagram

[244]

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:

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

Page 86, problem 325, solution diagram

[325]

Solution diagram gives 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]

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.

Page 198, problem 810, solution diagram

[810]

5 at a also works.

The solution from the book.

forces and after the points a and b are miai.

Note: This solution works even with outside liberties.

Page 204, problem 839, solution diagram

[839]

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]

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

• After , White b threatens to live with 6 points (including ), 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 (¿or should it be counted as a half-point ko?).

Page 214, problem 881, solution diagram

[881]

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

Page 218, problem 903, solution diagram

[903]

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.

Now a and b are miai.

Page 220, problem 910, solution diagram

[910]

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

Page 224, problem 929, solution diagram

[929]

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

Page 226, problem 937, solution diagram

[937]

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

Page 226, problem 938, solution diagram

[938]

3 at 4 also works.

Page 228, problem 945, solution diagram

[945]

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).

Page 230, problem 951, solution diagram

[951]

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.

Page 238, problem 986, solution text

[986]

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.