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03 Feb, 2006 Nightmare

 
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David Bryant
Gold Member
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Joined: 20 Jan 2006
Posts: 86
Location: Denver, Colorado

PostPosted: Fri Feb 03, 2006 6:04 pm    Post subject: 03 Feb, 2006 Nightmare Reply with quote

This is another "uniquitous" puzzle. After making the obvious moves, plus
some others that are not so obvious (X-Wing in the "8"s, XY-Wing rooted in
r6c6, a couple of connected pairs) I arrived at this position.

Code:
  79     6     1    28     5     4     3    79    28
  2     8    357   13    179   179   157    6     4
 379    4    357   123   178    6   15789 1579   27
 157    3     8    15    179    2    79     4     6
  4    57     6    58     3    789    2    579    1
 157    9     2    46    46    17    57     8     3
  6     2     9     7    48    13    48    13     5
  8    157   37    46    146   135   147    2     9
 35    157    4     9     2   1358    6    137   78


The puzzle is easily solved if we assume the solution is unique -- under
that assumption there must be a "1" at r8c5. And that is enough to
crack the rest wide open.

What if we don't want to make that assumption? One way to proceed works
fairly quickly,like this ("double-implication chain").

First, if we set r7c5 = 8 we will have the {1, 7, 9} triplet in r2c5, r3c5,
& r4c5, forcing the possibilities {4, 6} at r8c5, and guaranteeing that
any solution that can now be found will not be unique. So let's assume
that r7c5 = 8, and look for a contradiction.

A. r7c5 = 8 ==> "1" in r7c6, r8c6, or r9c6 ==> r6c6 = 7 ==> r2c6 = 9 ==> r5c6 = 8

B. r7c5 = 8 ==> r1c4 = 8 ==> r1c9 = 2 ==> r3c9 = 7 ==> r1c8 = 9

But now it's impossible to place a "9" in row 5. So r7c5 <> 8, and we
must have r7c5 = 4. And this is enough to solve the puzzle. dcb
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