View previous topic :: View next topic 
Author 
Message 
keith
Joined: 19 Sep 2005 Posts: 3315 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 

Back to top 


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 

Back to top 


Marty R.
Joined: 12 Feb 2006 Posts: 5732 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. 

Back to top 


keith
Joined: 19 Sep 2005 Posts: 3315 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 

Back to top 




You cannot post new topics in this forum You cannot reply to topics in this forum You cannot edit your posts in this forum You cannot delete your posts in this forum You cannot vote in polls in this forum

Powered by phpBB © 2001, 2005 phpBB Group
