The New Sudoku Players' Forum

[shakers]

Having solved March 11's Fiendish to my great surprise, I went back to look at March 4's fiendish, but having spent the past 90 minutes checking my pencil marks I seem to have reached an dead end.

Code: Select all

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


I sorely hope that I'm not missing something stupid, and there's a method that I'm overlooking...

[Pappocom]

Here are a few hints to get you re-started:

1. You should be able to place the 5 in box 1. Don't forget the upward thrust of the 5 in box 8.

2. Once you've done that you should be able to place the number in r2c3. To do so, you will have to understand the effect of 2 and 4 on box 3.

Those hints are deliberately cryptic, so that you still have some thinking to do!

- Wayne

[shakers]

Here are a few hints to get you re-started:

1. You should be able to place the 5 in box 1. Don't forget the upward thrust of the 5 in box 8.

5 is now placed in r2c2, and the effect of the 5 in box 8 is understood.

2. Once you've done that you should be able to place the number in r2c3. To do so, you will have to understand the effect of 2 and 4 on box 3.

The 2 and 4 have to go in either r1c8 or r3c7, but I cannot see how either can be tied to a particular cell; in turn I'm feeling very slow, but either cannot see the bearing of this information on r2c3, or do not fully understand the effect of 2 and 4 on box 3.

Those hints are deliberately cryptic, so that you still have some thinking to do!

...you're not kidding! ;)

[Pappocom]

That's good, you know where the 2 and 4 go in box 3. Don't worry about pinning them down to their actual cells just yet. You don't need to swallow the candy all at once!

The real significance is this: for box 3, you now know two cells where the other numbers cannot go.

To place the number in r2c3, you have to eliminate all the candidates bar one. At the moment, I guess it seems that 1 is still a candidate. But where does the 1 go in box 3? And there's another candidate you will need to eliminate, using the same principle.

I'm still being a bit obscure! Come back if you want a further hint.

- Wayne

[shakers]

That's good, you know where the 2 and 4 go in box 3. Don't worry about pinning them down to their actual cells just yet. You don't need to swallow the candy all at once!

The real significance is this: for box 3, you now know two cells where the other numbers cannot go.

To place the number in r2c3, you have to eliminate all the candidates bar one. At the moment, I guess it seems that 1 is still a candidate. But where does the 1 go in box 3? And there's another candidate you will need to eliminate, using the same principle.

I'm still being a bit obscure! Come back if you want a further hint.

Thanks very much for taking the time to point me in the right direction (twice!) I have now completed it... and I just hope that I'll be able to employ the relevant logic again in future puzzles!

Thanks again - it's really appreciated.