Comments  Talk
... by: Kwesi
Hi Andrew Enjoying your book immensely, and have now reached Puzzle 15 p94. The clue given was that there was an AIC, which I failed to find, I put this into your convenient solver after hours head scratching, and at the point on which I was struggling I came across a Unique Rectangle type 2. Since I had not come across one of these before, I read all about them in your book. and felt that I had understood them, and so returned back to puzzle 15 for a further head scratching session. D1&E1 are the floor candidates and contain 4&9 the cells D7 & E7 contain respectively 459 & 46789 to avoid the dreaded pattern at least one candidate 4 or 9 must be removed from either cell E7 or D7 , now I cannot understand why 9 can be removed at E7. All the examples only show 3 candidates at the most in one cell , so how do we know which of the dreaded candidates must be removed.
Thanking you once again for the excellent book and hoping that you can provide me with enough information to bring me back to normality. Kwesi
... by: Thos
Puzzle 3, page 23, used to illustrate Hidden Triples has an alternative solution that instead uses a Naked Quad in column 2.
Andrew Stuart writes: The exact order and way these early strategies are searched for has been tinkered with over the years so some examples in the book may solve in a different way. There are/were numerous reasons to update the approach over time. On triples/quads etc I think the main one was stopping one pair/triple/quad generating another in the same step. I was knock on effects to be separate steps so even though several or many pairs/triples/quads can be found in one step they should be independent of each other. That was not true when the book was compiled
... by: Thos
In case of the Exercise 5 discussed above there are further discrepancies between the diagram in the book and the display shown here. For example, the cells C3 and C7 are not shaded (coloured) in the book and cells F1, H2 and H9 are shaded in the book and not in the updated diagram. Also candidates 9 in cells F3 and G9 above are omitted in the book.
Also, in the diagram for Exercise 2 on p. 19 some candidate elimination has already been performed, e.g. 6 in F6 by the pair in A6, C6 and 7 in B7 by the pair in F7, H7.
... by: Bizman
On your resources list of puzzles above you have two number 26, but no number 27. Obviously the second one should be 27.
... by: Tony Minchin
Andrew Stuart  In your excellent book, at 12 SwordFish: P12 'dependant' (noun) should be 'dependent' (adjective). P36 'This clears all [the other] 6s in column 9 ...' P38 'In Fig 12.3 the sudoku board has, in total, three 5s in three rows..' ?in total, six 5s? P38 fourth para  '1, 5 and 8 will [have] 5 as the solution..' p39 The rule 'When there are: only three possible cells for a value..' But p38 at bottom said it could be 2 cells, 3 cells, 2 cells (not just 3,3,3) . And could it also be 3,3,2?
Andrew Stuart writes: Thanks for the addendum notes. If there is a new edition I'll be starting with new puzzles and examples and new typos. I did get the book proofed by a very profession person who did a great job. But then the publisher said I could have 50 more pages for no extra cost, so I got to write more and put more in  hence the typos in some sections. Appreciate the list and glad you are enjoying the book
... by: erik niebuhr
Dear Andrew Stuart, I am a beginner in Sudoku, just got your book (logic of S, exellent) but I have a problem with Exercise 2 (Quad, p19): In your answer (p209) you write that the quad of 2/6/7/9 removes the 2 in H2, there is no 2 in H2, it must be the 2 in H1, which means, that 2 should be placed in H3 ?? and thus no quad in row H. By the way, assuming the quad in column 1, I would remove 2/6/7/9 in B1 (and not only 2/7/9 as you suggest), together with 2 in H2 and 6/9 in J1, correct ?? Kind regards erik (denmark)
... by: Tom Ledwidge
Regarding the simple chain/colouring method I use + or instead of two different colours to avoid having coloured pens available.
I received your excellent book today and read it cover to cover. Tom
... by: Marwan Gharaibeh
Dear Mr. Stuart,,
Hi. I had your excellent book "The logic of Sudoku" and I am proud to say that in addition to the solver on the internet it has helped me reach the highest levels of solving. now the only puzzles I look for are the diabolical, sadistic and develish.
In page 100 , fig 29.3 of your book you talked about the ALS in the figure, but you didn't deal with (2) as a common restricted number for the two sets. in my understanding I see it as common restricted and I don't beleive that your definition applies to it, so why remove the (2)'s from A6 and C6?
May I bother you to respond to this please.
sincerely yours.
Marwan Gharaibeh United Arab Emirates
Andrew Stuart writes: Hi Marwan In Figure 29.3, 2 is not restricted common. It is present in A4 and C4. Only 1 is restricted common since 1 is the only candidate with the property that is is (a) present is both sets and (b) all occurances can be seen by both sets. Therefore it will be used in one or other set but not both. The 2 in B4 qualifies, but is disqualified by the 2s in A4 and C4  since these cannot see the other ALS in B6. Hope that helps
... by: Richard Kretzschmar
Concerning the Exercise 6 AnswerMultiColouring, how did you determine the shade sequence for the second chain? Assuming the shades for the 4 cell chain on the left, what is the logic for determining the shade, i.e. + or , for the 3 cell chain on the right? What would the solution be if the + and  were reversed in the 3 cell chain? Bottom line, is there a logical sequence for deciding where to start the second chain and with what symbol (or shade)?
Thanks.
Andrew Stuart writes: Hi Richard
There is a correct way of allocating the + or . Of the three cells in the second chain two can 'see' the other chains. These are the + cells, which in this case are E8 and J7. G8, the '' cell can't see any cell in the first chain. To rephase the rule  if some A+ and B+ cells can see each other (which they can and we thus lable them '+') and the A and B cells can't see each other (which they can't  and thus we lable them '') then all N that can see both A and B can be removed.
I know at first site it seems chicken and egg, but you looking out for what parts of the chain are visible to parts of other chains. 
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