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"Nightmare" for January 3, 2006

 
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keith



Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
PostPosted: Sat Jan 28, 2006 9:40 pm    Post subject: "Nightmare" for January 3, 2006 Reply with quote
Here is the Nightmare puzzle from January 3. I eventually solved it by identifying a few "unique rectangles" of different types.

Code:


Puzzle: DSN010306
+-------+-------+-------+
| . . . | . . 3 | 6 7 . |
| . . . | 6 . . | . . 9 |
| . . 8 | . 7 . | . . . |
+-------+-------+-------+
| . 5 . | . . 1 | . 8 . |
| . 7 6 | . 2 . | 5 3 . |
| . 2 . | 5 . . | . 9 . |
+-------+-------+-------+
| . . . | . 4 . | 2 . . |
| 1 . . | . . 5 | . . . |
| . 6 4 | 2 . . | . . . |
+-------+-------+-------+



If you do not use unique rectangles, I think this would be a very difficult puzzle indeed.

Go for it!

Keith

(I'll post my solution later.)
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David Bryant



Joined: 29 Jul 2005
Posts: 559
Location: Denver, Colorado
PostPosted: Sun Jan 29, 2006 11:39 pm    Post subject: Here's one way to get it done ... Reply with quote
Hi, Keith!

I've noticed that people over on the "Nightmare" site are calling this one Ruud's toughest puzzle yet. I did find a fairly simple way to crack it without relying on unique rectangles.

Using only the standard techniques I arrived at this position:
Code:
  2    149   59    149   159    3     6     7     8
3457   134   357    6    158   248   13    24     9
  6   1349    8    149    7    249   13    24     5
 349    5    39    349    6     1     7     8     2
 489    7     6    489    2    489    5     3     1
 38     2     1     5    38     7     4     9     6
3579   389  3579  1389    4     6     2    15    37
  1    389    2     7    389    5    89     6     4
 357    6     4     2    13    89    89    15    37

I worked on this long enough to realize that just fixing a single value somewhere wasn't enough to crack the puzzle wide open. So then I started looking for one more hidden pair.

My attention was drawn to a peculiar form of symmetry in the bottom left 3x3 box. Suppose that the value "9" is placed at either r7c2 or r8c2. Then the {3, 5, 7} triplet is revealed in the bottom left 3x3 box, the {1, 3, 4} triplet is created in the top left 3x3 box, the {5, 7} pair is uncovered in row 2, and r1c3 = 9.

Armed with this information we easily see that r7c2 <> 9 and also that r8c2 <> 9:

r7c2 = 9 ==> r1c3 = 9
r7c2 = 9 ==> r8c2 = 8 ==> r8c7 = 9 ==> r8c5 = 3 ==> r6c5 = 8
But now we have the {1, 5} pair in r1c5 and r2c5, leaving no possible way to complete r9c5, which must be a "1" because of r8c5 = 3.

The argument when r8c2 = 9 is almost exactly the same:

r8c2 = 9 ==> r1c3 = 9
r8c2 = 9 ==> r8c7 = 8 ==> r8c5 = 3 ==> r6c5 = 8
But now we have the {1, 5} pair in r1c5 and r2c5, leaving no possible way to complete r9c5, which must be a "1" because of r8c5 = 3.

So we can eliminate "9" from r7c2 & r8c2, leaving the {3, 8} pair in those two cells ... the rest of the puzzle is easily solved. dcb
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