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  'The Logic of Sudoku' Resources Page

Introduction
Welcome to the "Logic of Sudoku" resource page. This page is now being hosted on Scanraid because the orginal page on sudoku.org.uk has had the exercises and links to my solver removed.

To start with I have listed the exercises in the book with links to printable versions - in case you'd like to print them out instead of working in the book.

The second set of link are to my Step by Step solver for some of the puzzles and exercises in the book. This, I hope, will allow you to see the examples as the come about. In some cases some strategies might need to be turned off to allow the example to find the use of a particular strategy, but in most cases I have picked examples for the book that are necessary steps.

Any feedback or comments always appreciated.

Andrew Stuart

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Printable Exercises The following are the exercise puzzles in the book. You can click on the links to get a printable version of each puzzle.



The following are the sudoku puzzles in the book. You can click on the links to get a printable version of each puzzle or load the puzzle into Andrew Stuarts Step-by-Step solver

breakline Corrections
Hidden Pairs - page 20
The sentence immediately under the diagram should read "1 and 5 occur in only two cells in box 2, but they are hidden by the 3 and 8 in C4 and the 3/7/8 in C6"

Exercise 3 - Hidden Pairs
There is a small error on the exercise answer. One of the Hidden Pairs has been incorrectly highlighted. The fifth one, the {3/9} in box six is the correct Hidden Pair, not the {3/9} in box 8.
Correction to Exercise 3

Figure 22.2 on page 63
Unfortunately Figure 12.2 was substituted in place of the actual figure for 22.2. The real diagram is:

Figure 22.2

Table on page 123
The Extended Aligned Pair table on page 123 doesn't contain the strikethroughs which remove certain pair combinations and show how 3 can be removed. The corrected table is:
Table p123
Table on page 183
In the chapter 44, Killer strategies, is the list of cages with a unique combination of numbers. The correct table is here:
Unique Killer Cages
Note: Cages of size 7 were not included since they do not appear in most Killer Sudoku puzzles, but
41 = 2/4/5/6/7/8/9
42 = 3/4/5/6/7/8/9
A full list of all combinations is here: /Killer_Combinations.

If anyone wishes to comment, correct or contribute to these pages please feel free to contact me at andrew@sudokuwiki.org. I'm always interested to hear from other sudoku fanatics.

Andrew Stuart

Exercise 5 - Simple Chain / Colouring

I've received some emails about the difference between the solver and the solution published as exercise 5 in the book. This diagram to the right is currently how the solver sees the situation.

The book shows the elimination at G2 which the solver also shows and I have drawn with blue lines. What is missing from the book solution is the other two eliminations based on different chains. But it must be remembered that Simple Chains and Simple Colouring are two sides to the same coin. The solver uses Colouring to get rid of as many candidates as possible in a single move - but each elimination can be justified by a different chain. The red and green eliminations (and the blue) occur because the 9 can see cells with different colours.

There is a problem with the solution to exercise 5 in that I have only shown one elimination out of a possible three using chains only. That I believe arose because my solver was a more primitive version at the time of writing the book and it wasn't giving me the full picture.

 Exercise 5
Exercise 5: Load Example

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Comments...

Tuesday 15-Jun-2010

... 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)

Tuesday 25-Aug-2009

... 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

Sunday 19-Jul-2009

... 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

Wednesday 3-Jun-2009

... 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.


Monday 1-Jun-2009

... 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.

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Article created on 13-April-2008. Views: 4153
This page was last modified on 5-July-2009, at 02:02.
All text is copyright and for personal use only but may be reproduced with the permission of the author.
Copyright Andrew Stuart @ Scanraid Ltd, 2009