The New Sudoku Players' Forum

[kratos]

I'm sorry to revisit this one, but I have been stuck for some time and only just found this board. Funnily enough, I stuck at a point a bit further on than the other posters in the first discussion of this. I don't know whether to feel smug that I got through that, or very humbled at sticking at a point that nobody else has had problems with.
I have reached this:

4 6 * | 1 * 8 | * 2 *
* * * | 4 7 6 | 1 * *
* * 1 | 9 * 2 | 4 * *
-----------------------
8 * * | 5 * 4 | * * 1
* * * | * 1 7 | * * 4
6 1 4 | 8 9 3 | 7 5 2
-----------------------
2 8 6 | 7 4 9 | 5 1 3
1 4 * | 3 8 5 | 2 * *
* 3 * | * * 1 | * 4 *

The choices are such that I can see no way of deciding logically on any of the still blank cells.

What am I missing, please, anyone?

[shakers]

You know what r2c3 is, which will also give you r2c2.

You can work out that r4c7 and r5c7 are the only possibly places for a certain value, which rules the value out of r4c8 and r5c8.

This in turn gives you a pair between r2c8 and r4c8. Eliminating this pair from the rest of c8 gives you a value for r5c8, which in turns gives you r9c7.

Hopefully this should be enough to kick start the puzzle again for you.

[kratos]

Thank you for your help, but you are still ahead of me. r2c3 could be any of 2,3,5,8,9 according to me. r4c7 and r5c7 must contain a 6 between them, so I follow you as far as that goes, but r4c8 and r5c8 even without a 6 still contain 3,9 and 3,8,9 respectively. The 3's cannot be rulled out as the 8 still in r5c7 stops that. I am obviously missing something fundamental here, but I have done most of the other fiendish problems without too much trouble . This one is something else, and I don't know why.
Very frustrating for me, and probably for you trying to help

[shakers]

Thank you for your help, but you are still ahead of me. r2c3 could be any of 2,3,5,8,9 according to me.

True, but one of those numbers you have as possibles in r2c3 is the only occurance of it in box 1 which fixes it.

Having put this value into a r2 cell, a possible value in r2c2 is the only one in r2, again fixing it.

[su_doku]

Kratos

Slice and dice the fat lady for starters

[shakers]

r4c7 and r5c7 must contain a 6 between them, so I follow you as far as that goes, but r4c8 and r5c8 even without a 6 still contain 3,9 and 3,8,9 respectively. The 3's cannot be rulled out as the 8 still in r5c7 stops that.

OK... something's not quite right... so I'll talk you through everything I'm doing from a full set of pencil marks - you need to eliminate the 6 pencilmark from r9c7. You can do this because 6 must appear in r9 in box 8, so it can be eliminated from r9 in box 9.

This now means that the 6 in c7 must appear in box 4 (either r4c7 or r5c7), so it can be eliminated from r4c8 and r5c8.

The 3/9 pair between r2c8 and r4c8 means that the 3s and 9s can be eliminated from r3c8 and r5c8 and r8c8. As there's also a 3/9 pair between r1c7 and r2c8, the 9s can be eliminated from r1c9 and r2c9.

This elimination should give you r2c9 and r5c8. From there you should be away again.

[kratos]

Thank you for your patience, Shakers. I have now solved it

The clue came in your first of your two recent replies. I then found I had omitted to delete an 8 after filling in an 8 lower down. Once I had noticed this, the rest was easy. (Well as easy as any fiendish one).

[shakers]

Thank you for your patience, Shakers. I have now solved it

The clue came in your first of your two recent replies. I then found I had omitted to delete an 8 after filling in an 8 lower down. Once I had noticed this, the rest was easy. (Well as easy as any fiendish one).

Excellent - well done. It is far more satisfying to understand why they work, and not just be happy with looking up the answer.

I've recently begun to wonder whether an excess of pencilmarks is as helpful as it seems at first. Little slips, like yours, can lead to lots of heartache... Having said that, I think it'll take me a few years yet before I can attempt to solve them without *any* pencilmarks!