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keith
Joined: 19 Sep 2005 Posts: 3150 Location: near Detroit, Michigan, USA

Posted: Sat Mar 18, 2006 2:49 pm Post subject: DB Saturday Puzzle (March 18) 


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

Posted: Sun Mar 19, 2006 3:08 pm Post subject: I found a "6star constellation" 


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 "6star 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: 5126 Location: Rochester, NY, USA

Posted: Sun Mar 19, 2006 5:12 pm Post subject: 


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: 3150 Location: near Detroit, Michigan, USA

Posted: Sun Mar 19, 2006 6:24 pm Post subject: My solution 


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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