The New Sudoku Players' Forum

[LMC964]

Hi, new to forum, been doing Sudoku regularly (as opposed to the one in the Metro only if I run out of reading!) for about two months. Having completed Pete Sinden's Black Belt book, I bought the Times Fiendish and after no problems with #1-5 this one is driving me mad - what am I missing?

In desperation I resorted skipping and completing 7 & 8 to reassure myself that I might be able to do the Fiendish ones. Yep. So back to number 6 (posted below), and resorted to pencil marks - which I haven't used for AGES - but still no clue

Code: Select all

6    348 2378 | 348  5    234  | 37   1    9
38   9   1    | 368  36   7    | 5    2    4
347  5   237  | 349  1    2349 | 6    37   8
---------------------------------------------
2    1   4    | 67   67   8    | 9    5    3
389  7   389  | 349  2    5    | 48   6    1
389  6   5    | 1    349  349  | 2    48   7
---------------------------------------------
1    34  379  | 5    3479 6    | 3478 3478 2
5    348 378  | 2    347  34   | 1    9    6
3479 2   6    | 3479 8    1    | 347  347  5


Might it be an Ariadne's thread? Or is there some other technique I need to get to grips with...

Thanks in advance for any help

[re'born]

Look for a naked triple in column 1. This should be all you need to solve the puzzle.

[LMC964]

Sorry, I really don't see it. Would appreciate clarification please? - I've now solved the puzzle - by means of the hidden pair (4 & 7) in c1, which I am embarrassed that I didn't spot earlier :blush: (so thanks for making me examine that column more closely, even though I didn't 'get' your tip)

Hope I got the terminology right, still getting to grips with the names for all these principles.

[re'born]

Sorry, I really don't see it. Would appreciate clarification please? - I've now solved the puzzle - by means of the hidden pair (4 & 7) in c1, which I am embarrassed that I didn't spot earlier :blush: (so thanks for making me examine that column more closely, even though I didn't 'get' your tip)

Hope I got the terminology right, still getting to grips with the names for all these principles.

Well, you didn't 'get' the tip, but you found the exact same deduction (though for a different, but related, reason). You noted that 4 and 7 can only occur in the 3rd and 9th rows of column 1. Hence, r3c1=[47], r9c1=[47]. A different way to make this same deduction is to note that r2c1=[38], r5c1=[389], r6c1=[389]. These 3 cells contain a total of 3 different candidates. Thus, each must contain one of 3, 8 and 9. Therefore no other cell in column 1 can be 3,8 or 9, i.e., r3c1 <> 3, r9c1 <> 3, r9c1 <> 9. After this, you are left in the same position as after your hidden double.

I hope this helps. Did you check out the link from my first post?

[LMC964]

I did check the link - thanks, clear explanation on that site and have been using those principles for some time - I had a real thick moment on that puzzle lol.

Lesson learned from that puzzle,and #19 in the same book, which was also quite tricksy (and now solved) is that pencil marks sometimes muddy the waters rather than clarifying them - making it more difficult to focus on the numbers already fixed. It's like information overload I guess (sorry, I know there is a thread on this somewhere, I'll shut up now and go and read it again).