## Help with a Very Challenging puzzle

Advanced methods and approaches for solving Sudoku puzzles

### Help with a Very Challenging puzzle

I am new to Sudoku and find that the Easy, Medium, and most of the Hard I am able to do. I ventured into the Very Challenging section of the book I bought. I managed to do the first two but I'm having trouble figuring out what solving technique I should use on this one from here. I really don't want the solution, just the next step I should look for.

3 6 9 1 . . . 8 .
1 4 7 . 8 5 2 . .
5 8 2 . . . 1 . 4
7 . 6 5 . . 3 2 .
4 2 5 . . . . 1 .
8 . 3 . . 6 4 5 .
. 3 4 . . 1 . . .
. 7 1 8 5 . . 4 .
. 5 8 4 9 . . . 1

Candidates:

3 6 9 | 1 24 24 | 57 8 57
1 4 7 | 369 8 5 | 2 369 369
5 8 2 | 3679 367 79 | 1 369 4
-----------------------------------------------------------
7 19 6 | 5 14 489 | 3 2 89
4 2 5 | 379 37 789 | 6789 1 6789
8 19 3 | 279 127 6 | 4 5 79
-----------------------------------------------------------
29 3 4 | 267 267 1 | 5789 79 25789
269 7 1 | 8 5 23 | 69 4 2369
26 5 8 | 4 9 237 | 67 367 1
Last edited by Camchase on Sat Jan 07, 2006 3:43 pm, edited 1 time in total.
Camchase

Posts: 30
Joined: 03 January 2006

I was reading through some other posts and found this posted by Tso (in my language ) and tried it but it didn't work. I applied the number 2 into R7,8,9C1 in my puzzle using this strategy. Am I wrong?

I'm so confused

Thanks

For those who just want to use the tactic in it's basic form, here's all you need to know in layperson's English:

If your Sudoku has ONE cell with THREE candidates and all other undecided cells have TWO candidates -- you can immediately fill that three-candidate cell. Just check which candidate appears THREE TIMES in the row, column or box in which this three-candidate cell resides. That candidate is the one that goes in the three candidate cell.

For example, the following puzzle takes nothing but singles to go from the opening position ...

Code:
2 9 8 | . . . | 3 . .
. 4 . | . 1 3 | . . .
1 . . | . . . | 2 . .
-------+-------+------
. . . | 6 3 . | 7 . .
. . 9 | . . . | 8 . .
. . 3 | . 4 2 | . . .
-------+-------+------
. . 2 | . . . | . . 7
. . . | 5 9 . | . 6 .
. . 7 | . . . | 5 8 1

... to here ...

Code:
2 9 8 | . . . | 3 1 .
7 4 5 | 2 1 3 | 6 9 8
1 3 6 | . . 9 | 2 7 .
-------+-------+------
5 1 4 | 6 3 8 | 7 2 9
6 2 9 | . . . | 8 4 3
8 7 3 | 9 4 2 | 1 5 6
-------+-------+------
4 5 2 | . . . | 9 3 7
3 8 1 | 5 9 7 | 4 6 2
9 6 7 | 3 2 4 | 5 8 1

Code:
2 9 8 | 47 567 56 | 3 1 45
7 4 5 | 2 1 3 | 6 9 8
1 3 6 | 48 58 9 | 2 7 45
----------------+----------------+----------------
5 1 4 | 6 3 8 | 7 2 9
6 2 9 | 17 57 15 | 8 4 3
8 7 3 | 9 4 2 | 1 5 6
----------------+----------------+----------------
4 5 2 | 18 68 16 | 9 3 7
3 8 1 | 5 9 7 | 4 6 2
9 6 7 | 3 2 4 | 5 8 1

At this point, until very recently, most of us would solved this with an xy-wing or forcing chain. Instead, simply look at the row, column or box in which the lone three candidate cell sits. In each case, the candidate 5 is the only one that appears three times -- all others appearing twice or not at all. The BUG principle allows you to place the 5 in the r1c5.

Camchase

Posts: 30
Joined: 03 January 2006

Okay, I think I found a Hidden pair. Would the 5,8 be a hidden pair in R7C7,9 leaving 5,8 in both those cells and removing all the other candidates?

Thanks
Camchase

Posts: 30
Joined: 03 January 2006

Yes, 5,8 is a hidden pair in r7c7 and r7c9, or conversely 2,7,6.9 is a quad in r7c1,4,5 and 8.

Either way this leads to 2 being a hidden single in r8c9, which in turn leads to another quad of 5,7,8,9 in {r1c9,r4c9,r6c9,r7c9} and thus the elimination of 7,8,9 as candidates in r5c9. In turn we can solve r5c9=6, and similarly r2c9=3.

I have to leave now, so this is as far as I have gotten. Someone else will probably solve this before I get back, but if not, I'll check it out

Jeff
JeffInCA

Posts: 33
Joined: 02 January 2006

Thanks Jeff. I used the 5,8 hidden pair and, tada, solved the rest of the puzzle. Thank you so much for the confirmation on the hidden pair. I'm just starting to venture into the more difficult puzzles and some help is really appreciated. It's the first time I've used some of the more difficult solution techniques. I read the posts on this forum and I'm slowly starting to understand some of it. (Scary, isn't it ) But alot of it might as well be written in German for as much as I can understand it.

Thanks, again, for the help. I'm sure I 'll needs lots more in the future.

Cam
Camchase

Posts: 30
Joined: 03 January 2006

No problem Cam. I'm in the same boat as you. I just got started with Sudoko last month, and just found this site last weekend.

I'm glad I was able to help.

Jeff
JeffInCA

Posts: 33
Joined: 02 January 2006

Oh, by the way, the post you quoted before (by Tso) refers to a really new and advanced technique (one that I only partially understand), called the BUG principle.

What Tso was mentioning was what seems to be the simplest form of using this BUG principle, and the only version I DO understand.

But for this to work, you have to have your puzzle reduced to the point where all the remaining cells have only two candidates except for one cell which has 3. Then you use the logic Tso described to solve the tri-valued cell and hopefully the rest of your puzzle falls into place.

I haven't really had a chance to practice using this yet, but I'm looking forward to trying.

Your puzzle still had many 3 and 4-valued cells remaining so it wasn't yet at a state where this principle.

Caveat: I may have gotten part of this wrong, so definitely don't take it as gospel. If anyone out there reading this wants to correct me, feel free

Jeff
JeffInCA

Posts: 33
Joined: 02 January 2006