Solving the Nightmare . . .

[pepi]

I do not understand how I should be getting from

1249 12 6 7 1489 1249 35 48 35
49 8 5 39 6 349 7 1 2
1247 3 47 128 148 5 6 48 9
239 6 389 5 349 349 2348 7 1
1379 4 3789 13689 2 17369 38 5 36
5 12 378 1368 1348 17346 2348 9 346
6 7 2 4 5 19 19 3 8
8 9 1 236 7 236 54 26 54
34 5 34 26 19 8 19 26 7

to the solved puzzle

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

I succeeded by making some fuzzy-logic assumptions on picking the final numbers the pairs in R1C7/9 and R1C2/6, mostly based on which option in each case would force the most other choices. I also forced the 7 into R3C3. Is this akin to some of the rather ellaborate strategies detailed on the site?

How did other folks solve this?

After the first hour, I had only solved 5 cells. After the second hour, only 7 more. This was certainly a challenge. My first Nightmare took 6 hours and 5 minutes to solve (which included lunch and about 1/2 hour each of the Dukes of Hazard movie and Final Destination 3 -- not my rental).

All the best,

Pepi

[Ruud]

Hi Pepi,

I usually leave these types of questions to the real professionals that frequent this forum, but now I have the chance to welcome you and show you a nice trick.

This is the candidate grid that can be reached with basic techniques:

Code: Select all

.---------------.---------------.---------------.
| 1249 12   6   | 7    1489 1249| 35   48   35  |
| 49   8    5   | 39   6    349 | 7    1    2   |
| 1247 3    47  | 128  148  5   | 6    48   9   |
:---------------+---------------+---------------:
| 23   6    389 | 5    3489 49  | 2348 7    1   |
| 137  4    3789| 1689 2    1679| 38   5    36  |
| 5    12   378 | 168  1348 1467| 2348 9    346 |
:---------------+---------------+---------------:
| 6    7    2   | 4    5    19  | 19   3    8   |
| 8    9    1   |*236  7    236 | 45  *26   45  |
| 34   5    34  |*26   19   8   | 19  *26   7   |
'---------------'---------------'---------------'

I marked 4 cells in this grid with an asterisk.
These cells occupy 2 rows, 2 columns and 2 boxes.

They almost have 2 candidates each, digits 2 and 6.
If this were the case, the puzzle would have 2 solutions. You can test this for yourself.
The only candidate that prevents this from happening is candidate 3 in R9C4. This candidate must be placed in this cell to prevent the deadly multiple solutions pattern.

This technique, although founded in esotheric theory, is actually quite simple to spot and execute. It is called a Uniqueness test 1, also known as a Unique Corner.

You could have spotted the XYZ-Wing in box 2, but that wouldn't have gotten you far.

If you have moral objections to uniqueness based solving techniques, just say so and one of the professionals will no doubt show you another way to solve this Nightmare.

Enjoy the site,

Ruud

[David Bryant]

Hi, Pepi!

Here's another way to solve the puzzle from the position Ruud posted.

Code: Select all

.---------------.---------------.---------------.
| 1249 12   6   | 7    1489 1249| 35   48   35  |
| 49*  8    5   | 39   6    349 | 7    1    2   |
| 1247 3    47  | 128  148  5   | 6    48   9   |
:---------------+---------------+---------------:
| 23   6    389 | 5    3489 49* | 2348 7    1   |
| 137  4    3789| 1689 2    1679| 38   5    36  |
| 5    12   378 | 168  1348 1467| 2348 9    346 |
:---------------+---------------+---------------:
| 6    7    2   | 4    5    19  | 19   3    8   |
| 8    9    1   | 236  7    236 | 45   26   45  |
| 34   5    34  | 26   19*  8   | 19   26   7   |
'---------------'---------------'---------------'

Assume first that r2c1 = 9. Then we have

r2c1 = 9 ==> r2c4 = 3 ==> r2c6 = 4 ==> r4c6 = 9
r4c6 = 9 ==> r7c6 <> 9 ==> r9c5 = 9

See the three cells marked with an asterisk above. It's obvious that we cannot now possibly place the digit "9" anywhere in the top center 3x3 box. Therefore r2c1 = 9 is impossible, and we must have r2c1 = 4 (and the rest of the puzzle is easily solved from there). dcb

PS The method Ruud described makes the assumption that the solution to the puzzle is unique. I don't have any objection to making that assumption -- especially not with Ruud's excellent puzzles! But I usually search for another way around these "uniqueness patterns", just to make them more interesting.

PPS In looking at the grid you posted, Pepi, it appears that you need a little more practice with the "row/column on box interaction" technique. Here's the grid as you posted it.

Code: Select all

1249 12  6    7     1489  1249  35   48  35
49   8   5    39    6     349   7    1   2
1247 3   47   128   148   5     6    48  9
239  6   389  5     349   349   2348 7   1
1379 4   3789 13689 2     17369 38   5   36
5    12  378  1368  1348  17346 2348 9   346
6    7   2    4     5     19    19   3   8
8    9   1    236   7     236   54   26  54
34   5   34   26    19    8     19   26  7

If you look closely at column 3, you'll see that the "9" must fall in r4c3 or r5c3. In other words, the "9" in the middle left 3x3 box must fall in column 3. So you can eliminate "9" from r4c1 & r5c1,

Similarly, if you examine column 5 you'll see that the "3" in that column must fall in the middle center 3x3 box -- either at r4c5 or r6c5. So you can eliminate "3" from r5c4, r6c4, r4c6, r5c6, and r6c6.

After making those eliminations I think you'll find that your grid exactly matches the one that Ruud posted.