Note. I have read most of the book thoroughly and solved many of the puzzles by hand and the regular sudokus with my computer solver. It is a good piece of writing, but as with almost any book there are mistakes and sections that could be made better or more complete. "You" in the comments refers to the author (Stuart). I haven't bothered about most small typos. If anyone finds an error in this text, please let me know (sudoku@anderbok.com). The latest additions are shaded in green.

Chapter 1. Known and Unknown in Sudoku

1. Page 4, row 13. "Not including re-labelling, it has been shown that there are 3,359,232 symmetries. Many of these lead to the same grid, so ..."
   Comment 1. You should have shown the calculation, because it is very easy: rows within a "chute" (group of rows) can be permuted in 3! = 6 ways. There are 3 chutes for rows, and the chutes likewise can be swapped in 6 ways. The same goes for columns. Finally, columns and rows can be interchanged (transposed). So the total number of permutations (symmetries) is (6 * 6³)² * 2 = 6⁸ * 2.
   Comment 2. When you say "many of these lead to ..." I think that "many" is an exaggeration. Comparing the number of essentially different boards times the number of symmetries with the total number of completed boards gives much less than 1 % surplus (about 1 in 19,000): 5,472,730,538 * 362,880 * 3,359,232 / 6,670,903,752,021,072,936,960 = 1.000,051,630.

2. Page 15, row 2. "There are 38 candidates that can be removed ..."
   Correction. I found 42 eliminations. See Chapter 49, Exercise Answers, for the additional 4 ones.

3. Page 18, row above Figure 5.1. "8 from H8."
   Correction. "4 and 8"

4. Page 19, row 2 (Exercise 2). "There are two Naked Quads on this board."
   Comment. Actually, the first quad is in a column with seven unsolved cells, so it is also a hidden triple. The second quad is in a row with only five unsolved, so it is more easily described as a hidden single. See also Chapter 49, Exercise Answers.

5. Page 20, row 9. "... and the 1/3/7 in C6."
   Correction. "3/7/8"

6. Page 21, Figure 6.2. C4=[5/8], C6=[5/8].
   Correction. C4=[1/5], C6=[1/5]. This means that no further elimination can be done in row C at this stage.

7. Page 25, Figure 9.2. The figure is the same as Figure 10.2.

8. Page 25, row 2 from bottom. "Here there are three 3s in box 3 and aligned on row A, which tells us that we can remove them from the rest of the box."
   Correction. The text accurately discribes the above figure, but this is an example of Box/Line Reduction, and not a Pointing Triple.

9. Page 27, row 5 from bottom. "If any one number occurs twice or three times in just one unit, then we can remove that number from the intersection of another unit."
   Comment. Although the meaning of this sentence is clear from the subsequent summary, I have trouble with the grammar. Firstly it's not enough that a number occurs twice or three times in a unit, and secondly you don't eliminate from the intersection but from the cells outside the intersection. A better formulation would be e.g. "If a number occurs in only two or three cells in one unit, and these cells are all in its intersection with another unit, then we can remove that number from the rest of the other unit."

10. Page 28, last rows. "seven Pairs and three Triples"
    Correction. "one Triple"
    Comment. There are three possible triples, but two of them are covered by other intersection removals. Counting this way we would also have three additional pairs.

11. Page 34, row 3 from bottom. "there are at least two X-Wings"
    Comment. I found two or four X-wings, depending on the definition, but only one of them is necessary for the solution. There may be additional X-wings depending on the solving order. Some definitions require the cells A, B, C, D in Figure 11.1 to be in four different boxes, because if the cells are in only two boxes, the X-wing can be replaced by intersection removals.

12. Page 41, row 3. "If we know a solution is bound to exactly four rows and four columns and we know there must be a total of nine of that number to place on the board, then the remaining five rows and five columns also form a restricted set. ... In fact for every size – N fish there is a size – (9 – N) counterpart."
    Comment. It is not expressed clearly enough that this goes only when all the nine instances of a number are still unsolved. In the general case when U are unsolved, a size – N set has a size – (U – N) counterpart. Of course singles should be solved first, and a large set can sometimes be divided into more than two parts, e.g. U = 7 = 3 + 2 + 2.

13. Page 44. In Figure 16.2 one arrow is missing in the top row.

14. Page 47, Exercise 5. In the figure I don't understand the cells F3 and G9. My solver gives the result shown below, after reduction by the pair 1/9 in column 6. I can't find a method (of equal or less complexity than colouring) to eliminate 1/9 in F3 and 9 in G9. Also, if the figure in the book is correct, we would have an X-wing in C3, C7, G3, G7, which eliminates (among other candidates) 9 in G2. That makes the colouring method unnecessary in this case.
    

15. Page 47, row 3 from bottom. "... contains 12 links"
    Correction. "11 links".
    Comment. Because B2 and G7 have opposite colours, there must be an odd number of links between them. Also, the long chain isn't necessary. The chain B2-C3-C7-G7 contains only 3 links.

16. Page 54, Figure 18.4. This is a bad example, because it's more easily solved using the pair in C1/C3, which gives a naked single in A3.

17. Page 59. You could have mentioned that a WXY-wing (without Z in the pivot cell) is an extension of the XY-wing (Y-wing), just like the WXYZ is an extension of XYZ.

18. Page 61, last rows. "Find these two other Y-Wing Chains and the candidates that are eliminated. No additional candidate elimination is necessary from this point to identify the solutions."
    Correction. No, this is not enough. You need an XY chain from H1 to H9 to eliminate 9 in H5 (and H4 and H6), and then another chain from H5 to G8 to eliminate 5 in G5 (and H9). This makes the previous Y-wing chains redundant.

19. Page 63, Figure 22.2. Wrong figure. This is a copy of Figure 12.2!

20. Page 69, row 8 from bottom. "method of candidate ..."
    Comment. One word or one row seems to be missing.

21. Page 74, Figure 25.3. If it is supposed to be identical to Figure 25.4, which seems to be the case, there is one 9 missing in D1.

22. Page 76, Exercise 8.
    Comment. This is a very difficult sudoku, which requires repeated use of ALS or other complex techniques.

23. Page 86, row 4 from bottom. "This makes the links to the group from C8 and down to J7 strong links."
    Correction. No, the link from A7/B7/C7 to J7 is weak, because there is an 8 in E7 (and G7).

24. Page 87, row 2. "The elimination in Figure 26.4 immediately gives us another grouped X-Cycle on this number."
    Comment. No, this X-cycle is independent of the previous one, since the link in column 5 is weak.

25. Page 87, row 8. "Exercise Answers begin on Page 202."
    Correction. "Page 208"

26. Page 89, Figure 27.1.
    Comment. Four candidates have already been removed using techniques from later chapters. This is quite unnecessary because the chain works fine anyway.

27. Page 93, bottom row. "the 5 cannot be in B8 and the 9 cannot be in B9."
    Comment. Not the best example, because we already know 9 is not in B9.

28. Page 101, last rows. "The XZ Rule is required just once here. Apart from a Simple Chain, it should be easy sailing."
    Comment. As far as I can see, ALS is required twice (or some other advanced technique). First to eliminate 8 from G8 and H8, and then to eliminate 8 from J2.

29. Page 103, Sudoku Puzzle 17
    Comment. This one is really difficult – almost impossible to solve with AIC and ALS (took me several hours). But there are a few faster ways.
    (1) After solving singles and basic methods, we see in the figure below that some connected cells have 1/6. Using a uniqueness strategy, similar to the ones in part five of the book, we can try to set r8c2=6 and r7c3=1. This leads to 1/6 in r2c2/r2c3 and 1/6 in r7c5/r8c5, which is a forbidden pattern. Hence r8c2=1 and r7c3=6. If we then try to set r8c5=6, we mustn't have r2c2=6 to avoid the forbidden pattern. With r2c2=4 we get a contradiction pretty fast, and so r8c5=6 is wrong. Finally we use an xy-wing – marked in the right-most figure – to eliminate some 2:s, and the rest is easy.
       
    (2) This solving path is even easier, but more arbitrary, so it could be called guessing. After solving all the singles, we note some pairs 5/8. Setting r2c7=5 leads to a contradiction. Thus r2c7 must be 8, and the remaining cells can be solved with singles only.
       
    (3) If we insist on solving with ALS, here is one way. We can remove 2, 5 and 8 from the center cell r5c5, leaving only 6. If r5c5 is 2, it's not 6, so r5c1=6, r4c1=8 and r4c4=2, a contradiction in box 5. If r5c5 is 5, then r3c5=8, r3c9=5, r6c9=8. There is an ALS of 2/5/8 in r5c2/r5c8, which collapses to 2/5, a contradiction in row 5. If r5c5 is 8, then r3c5=5, r3c9=8, r6c9=5. There is an ALS of 2/5/8 in r4c4/r6c6, which collapses to 2/8, a contradiction in box 5. When we know that r5c5 is 6, we get the same situation as in the third figure above.
         

30. Page 110, row 7. "Figure 31.1"
    Correction. "Figure 32.1"

31. Page 112, last rows. "The General Pattern Exclusion Rule states that if a pattern prevents another number from having a valid solution pattern itself, then that pattern cannot be true. We'll discuss more of this later, after we've proven how Finned X-Wings work."
    Comment. I can't find any further discussion of the General Pattern Exclusion Rule, neither in this nor in following chapters.

32. Page 113, Figure 32.6. One of the 8s is misplaced.

33. Page 113. Figure 32.8 is missing.

34. Page 114. Figure 32.9 is identical to Figure 32.11. The cell with 'bfjn' should be shadowed.

35. Pages 122–123. The figures have wrong numbers. It should be Figure 34.4, 34.5 and 34.6.

36. Page 122, rows 5–9. "So why use APE at all? In the first example, the two cells targeted for reduction were both tri-value. Y-Wing works on bi-bi-bi cells, and XYZ-Wing on bi-tri-bi cells, so neither strategy is correct for this situation."
    Comment. There is a logical misconception here. Because J2 is tri-value, XYZ-Wing could have worked – had there been different numbers in the cells – and it doesn't matter how many candidates are present in the target cell H2. In this case a different extension of XY-wing works fine: WXY-wing with pivot cell J2=3/4/6 and pincer cells G1=2/4, C2=2/6 and D2=2/3, which eliminates 2 from H2.

37. Page 123, table in rows 5–10. 5/6 is a visible pair in C8. In the right column of the table, row 1 (3 and 5), row 2 (3 and 8), row 3 (6 and 5) and row 6 (8 and 8) should be struck out. This makes it possible to eliminate 3 from C2.

38. Page 150, figure "Option A for 6". "If it's in F2, then 6 is forced into D5"
    Correction. No, 6s may also be in D4/H5. The conclusion that F2 cannot contain a 9 still holds, though.

39. Page 156, Figure 41.1. This puzzle actually has three solutions, one for each E5=[2/4/7]. So BUG is not applicable.

40. Page 172, paragraph 5. "We know that all the numbers in a row, a column or a box always add up to 45, since the numbers 1 to 9 occur once in each unit. If we have a box that is partially within the region, it will contain a certain set of numbers outside the region. In order for the 45 rule to be satisfied for all boxes as well as rows and columns, any deficit must be made up elsewhere – with identical sets of numbers. With two rows or columns, the sum we need to have in mind is 45+45=90. Likewise for three rows or columns, the sum is 135."
    Comment. It isn't appropriate to talk about sums in sudoku, because – as you have stated in chapter 2 – the numbers are only symbols. And you do point out that the sets of numbers must be identical, which has no relation to sums.

41. Page 175, last paragraph. "We end up with B5 and D9 as a Naked Pair of 4/3. This is a conjugate pair ..."
    Comment. No B5=D9, not a naked pair (but D5 and D9 is). Also, as I understand the definition of conjugate pairs, the two cells must be in the same unit, which is not the case here.

42. Page 182, row 3 from bottom. "a triple cage starting in D7"
    Correction. "D4"

43. Page 182, last sentence. "Because we have a 4 on E3, we know that E4 and F4 can be only a 1 and a 2."
    Comment. No, 4 can only be eliminated from E4, in the same row as E3. 4 can't be eliminated from F4 until later, when we have solved F7.

44. Page 183, Figure 45.1. The cage in D7/D8 (2[7]), should not be shadowed, because it does not lead to a single set, but three different: 1/6, 2/5, 3/4.

45. Page 183, table at the bottom. The list is incomplete. Also, it shouldn't contain 2[5] (two combinations: 1/4, 2/3). It should look like this.

    Cages of 2	Cages of 3	Cages of 4	Cages of 5	Cages of 6
    3   1/2	6   1/2/3	10  1/2/3/4	15  1/2/3/4/5	21  1/2/3/4/5/6
    4   1/3	7   1/2/4	11  1/2/3/5	16  1/2/3/4/6	22  1/2/3/4/5/7
    16  7/9	23  6/8/9	29  5/7/8/9	34  4/6/7/8/9	38  3/5/6/7/8/9
    17  8/9	24  7/8/9	30  6/7/8/9	35  5/6/7/8/9	39  4/5/6/7/8/9

46. Page 184, row 6. "one cell of which is in box 6"
    Correction. "box 4"

47. Page 185, Figure 46.2. The 1 in G4 is in the wrong place (it is probably meant to be in G5, but is not necessary for a unique solution). Also, some of the sums are missing.

48. Page 208, row 1. "start on page 192."
    Correction. "page 220"

49. Page 208, row 4. "... remove 38 candidates"
    Correction. "42 candidates". The 4 extra eliminations are marked with yellow in the figure below (1 and 4 in B6, 5 and 9 in F1).
    

50. Page 209, Exercise 2. In the figure, 6 can be eliminated in B1, so it should be shadowed. In H3, 5 should be shadowed instead of 2.

51. Page 210. In the figure, one pair is erroneously circled. Instead of G4/J4, D9/F9 is a hidden pair of 3/9.

First published in 2009.
Last modifed 2011–10–18.