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DB Saturday Puzzle (March 18)

 
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keith



Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
PostPosted: Sat Mar 18, 2006 2:49 pm    Post subject: DB Saturday Puzzle (March 18) Reply with quote
Here is today's David Bodycombe puzzle:

Code:


Puzzle: DB031806 ******
+-------+-------+-------+
| 2 . 6 | . . . | . 9 . |
| . . 8 | . 9 . | 7 . 6 |
| . 9 . | 5 . . | . 1 . |
+-------+-------+-------+
| . . . | . 5 8 | . 3 . |
| . 7 . | . . . | . 8 . |
| . 2 . | 7 1 . | . . . |
+-------+-------+-------+
| . 8 . | . . 7 | . 5 . |
| 4 . 5 | . 8 . | 6 . . |
| . 6 . | . . . | 4 . 8 |
+-------+-------+-------+



My solution, which I do not think is very elegant, involves a contradiction and then coloring.

Keith
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David Bryant



Joined: 29 Jul 2005
Posts: 559
Location: Denver, Colorado
PostPosted: Sun Mar 19, 2006 3:08 pm    Post subject: I found a "6-star constellation" Reply with quote
Hi, Keith!

These Bodycombe puzzles are definitely getting more intricate. After making the "obvious" moves and eliminations I arrived at this position.
Code:
  2    14     6     8     7    14*   35     9    35
 13     5     8    123    9    123    7     4     6
 37     9    47     5    46    346    8     1     2
  6    14*   149   49     5     8     2     3     7
  5     7    349  3469    2   3469   19     8    149
  8     2    349    7     1    349   59     6    459
 19     8     2    46    46     7    139    5    139
  4     3     5    129    8    129    6     7    19
 179    6    17    19     3     5     4     2     8

The best I could do here is use the "6-star constellation" defined by r1c6 & r4c2.

A. These cells are linked through r1c2 -- they're either both "1" or both "4".
B. If they're both 4 then r3c5 = 6 ==> r3c6 = 3
C. But we also have r4c4 = 9 ==> r6c6 = 3

So both r1c6 & r4c2 must be "1" ... the rest is fairly simple.

I also noticed that the three cells r7c1, r8c9, and r9c4 are similarly linked -- either all three of them are "1", or all three are "9". I didn't see a convenient way to use that information, though. dcb
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Marty R.



Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
PostPosted: Sun Mar 19, 2006 5:12 pm    Post subject: Reply with quote
I used basic techniques, then one forcing chain solved about eight cells and opened the floodgates.
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keith



Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
PostPosted: Sun Mar 19, 2006 6:24 pm    Post subject: My solution Reply with quote
David,

Here is what I did:

From the position you posted, assume R9C1 is <1>, and follow the two implications:

R9C1 has been set to <1>.
R9C3 must be <7>.
R3C3 must be <4>.
R1C2 must be <1>.

R9C1 has been set to <1>.
R9C4 must be <9>.
R4C4 must be <4>.
R4C2 must be <1>, which is a contradiction in C2.

So, R9C1 is not <1>. Then, using coloring on <1>:

R9C4A R9C3a R7C1A R2C1a R2C4A. There are two squares labeled "A" in C4, thus the squares labeled "a" must be <1>, and all the remaining moves are forced.

Keith
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