... 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
... by: Andrew Stuart
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.
... by: Richard Kretzschmar
Concerning the Exercise 6 Answer-Multi-Colouring, 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.
