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Another good puzzle - 12/31

 
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keith



Joined: 19 Sep 2005
Posts: 3184
Location: near Detroit, Michigan, USA

PostPosted: Sat Dec 31, 2005 1:58 pm    Post subject: Another good puzzle - 12/31 Reply with quote

Try this one:

Code:

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



Happy New Year!

Keith
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David Bryant



Joined: 29 Jul 2005
Posts: 559
Location: Denver, Colorado

PostPosted: Sun Jan 01, 2006 12:12 am    Post subject: Here's one way to get there ... Reply with quote

This puzzle was in the Rocky Mountain News this morning.

The first 22 moves were fairly routine, leading to this position, with 29 squares unresolved.
Code:
   35    358    358     4      7      1      6      2      9
   9      4      6      3      58     2     178     57     15
   7      2      1      68     9     568     38     4      35
  356    3578   358    678     1      9      2      57     4
   56     9      2      67     3      4      17     8     156
   1      78     4      5      2      68     37     9      36
   4      1      35     2      58    358     9      6      7
   2      6      9      1      4      7      5      3      8
   8      35     7      9      6      35     4      1      2

From here I solved the puzzle by considering a "double-implication chain" rooted in r6c8:

r6c8 = 3 ==> r6c9 = 6 ==> r6c6 = 8
and
r6c8 = 3 ==> r3c8 = 8 ==> r3c4 = 6 ==> r5c4 = 7 ==> r4c4 = 8

But this is a contradiction (two "8"s in the middle center 3x3 box), so we must have r6c8 = 7, and the puzzle is easily solved from there. dcb
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keith



Joined: 19 Sep 2005
Posts: 3184
Location: near Detroit, Michigan, USA

PostPosted: Sun Jan 01, 2006 4:12 am    Post subject: A simpler chain? Reply with quote

David,

There is a chain which you may consider simpler:

From your position,

Assume R3C6 is <5>, then R3C9 = <3>, R6C9 = <6>, R6C6 = <8>, leading to the contradiction: R3C6 is pinned to be <6>.

So, R3C6 is not 5, the rest is trivial.

Happy New Year!

Keith
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