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keith
Joined: 19 Sep 2005 Posts: 3260 Location: near Detroit, Michigan, USA

Posted: Sat Feb 18, 2006 1:44 am Post subject: One more puzzle (DB021706) 


Here is today's David Bodycombe puzzle.
Code: 
Puzzle: DB021706
++++
 . . 5  . . .  . . 2 
 7 9 2  . 1 .  6 . . 
 . 6 .  . . .  . . . 
++++
 . 1 .  . . 3  . . 8 
 4 . .  8 . 9  . . 5 
 2 . .  7 . .  . 6 . 
++++
 . . .  . . .  . 4 . 
 . . 4  . 6 .  2 9 3 
 6 . .  . . .  1 . . 
++++

Last week's Friday puzzle (DB021706 in the thread "Two more Puzzles") is great training for this one!
Personally, I very much enjoy these.
Keith 

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Steve R
Joined: 24 Oct 2005 Posts: 289 Location: Birmingham, England

Posted: Sat Feb 18, 2006 3:30 am Post subject: David Bodycombe’s 17 February 


Keith
I enjoy them too.
I’m not an enthusiast of uniqueness assumptions and forcing chains so these puzzles, involving at most an XYWing, are just up my street.
The other aspect which appeals is that the grading seems reasonably consistent. You do not categorise this particular example but I should guess about five stars. Consistency prevents me from embarking on a trail that I cannot finish within an acceptable time.
Thank you for providing access to DB's weekend stuff.
Steve 

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Guest

Posted: Sat Feb 18, 2006 3:59 am Post subject: 


I believe by definition a valid Sudoku must have only one solution, so uniqueness assumptions are actually givens. If it has more than one solution, then logic will not and cannot solve it.
Tracy 

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keith
Joined: 19 Sep 2005 Posts: 3260 Location: near Detroit, Michigan, USA

Posted: Sat Feb 18, 2006 6:50 am Post subject: DB's puzzles 


WARNING: SOLUTION HINTS CONTAINED BELOW!!
Steve,
This one is five stars. I agree that the difficulty and the required solution techniques are just about right (for me).
Tracy,
You are correct. The implied contract in a Sudoku puzzle is: Find THE solution to ...
Even if the contract were: Find ONE solution to ..., uniqueness techniques would apply. They only may fail if the contract is: Find ALL solutions to ...
David Bryant has a great (serendipitous) example of a puzzle that has a few multiple solutions. I found that even if you use "uniqueness" techniques, the fact of multiple solutions is obvious.
That said, my own belief is: If published puzzles were to have multiple solutions, I would soon lose interest. I'm with Steve: Bring on the wings and the fishes, but I kind of draw the line at chains, at least until someone can give a coherent explanation of a systematic way to find them. (That I, a mere mortal, can understand and use. In my opinion, David Bryant and someone_somewhere are not mere mortals, they are geniuses. I cannot fathom how they see this stuff)
Anyway, the current puzzle is interesting. What I found is that there are two unique rectangles that eliminate some possibilities, but they are not needed. Even if you find them, you still need to find an XYwing in the top right area of the puzzle. Then, like last week's puzzle, the final step is either a BUG (which assumes uniqueness), or (another) XYwing (which does not).
Happy puzzling, (and thank you for the soapbox)
Keith 

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TKiel Guest

Posted: Sat Feb 18, 2006 2:25 pm Post subject: 


Keith,
Although I too only like to do puzzles with a single solution, I know some people who do puzzles with multiple solutions because they get many puzzles for the price of one and for the challenge of finding ALL the possible solutions for a particular puzzle. Maybe those kinds of puzzles need to have a different name.
Quote:  Bring on the wings and the fishes, but I kind of draw the line at chains 
The proof behind all the wings and fishes is that they are chains, so it you use them you do use chains. It's just that they are fairly easy to spot and generally short enough that the exclusion is apparent, even for a paper and pencil solver. I'm sort of with you on what most people call forcing chains (for example, where assigning either candidate in a certain bivalue cell implies that a different cell must be a 6) because I have not seen an explanation that was logical for why that particular starting cell was chosen. I don't discount the logic behind the technique, but there seems to be no logical way for deciding where to start.
However, I do use chains that are conjugate links (up to three different ones can be linked to make exclusions) because the logic behind it and the logic of where to start is apparent. And it can be done in your head, although using helper software makes it much easier. They've enabled me to solve more advanced puzzles that don't fall to the more familiar patterned chains like swordfish, xwing, etc... 

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

Posted: Sat Feb 18, 2006 11:29 pm Post subject: Where to start looking ... 


Tracy wrote:  ... I have not seen an explanation that was logical for why that particular starting cell was chosen. 
I like the "doubleimplication chain" technique. It's not always easy to find the right cell to start from, but I have developed a couple of heuristic rules. Here's an example  it's from Ruud ven der Werf's Nightmare puzzle for today, February 18.
After making the obvious moves I arrived at this position.
Code:  26 9 1 78 4 78 36 5 236
235 8 235 9 6 15 4 27 127
4 7 56 3 2 15 18 68 9
8 6 7 5 1 3 2 9 4
123 23 9 468 7 468 5 368 168
135 35 4 2 9 68 18 3678 1678
7 235 235 1 8 9 36 4 2356
236 4 8 76 5 276 9 1 23
9 1 256 46 3 246 7 28 258 
I can see a "nonunique rectangle" starting to form at r1c7, r1c9, r7c7, & r7c9. I can be sure that there's either a "2" at r1c9, or else there's a "2" or a "5" at r7c9. I don't see how to use that information directly, but I can use it to help find a "doubleimplication chain."
The heuristic rule I've developed says that any cell that's conjugate to a cell like r1c9 or r7c9 in this situation is a good starting point for a chain. I'm not sure why that is, but I've found it to be true in several dozen very tough puzzles, so I have faith in the rule.
Anyway, in this case my heuristic rule tells me to look for cells containing a "2" that's conjugate to r1c9 or r7c9, or for cells containing a "5" conjugate to r7c9. Here are some results.
A. r2c8 = 2 ==> r9c8 = 8
B. r2c8 = 2 ==> {3, 6} pair in r1c7&c9 ==> r3c8 = 8
Therefore r2c8 <> 2 and r2c8 = 7 ... this solves the puzzle.
C. r8c9 = 2 ==> r9c8 = 8
D. r8c9 = 2 ==> {3, 6} pair in r1c7&c9 ==> r3c8 = 8
Therefore r8c9 <> 2 and r8c9 = 3 ... this is also enough to solve the puzzle, although it's more work this way than if you find the "7" above.
I didn't find any good cells involving a "5". And a couple of the cells involving the "2" were duds. But this simple rule did help me uncover two closely related chains that solve the puzzle. dcb 

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keith
Joined: 19 Sep 2005 Posts: 3260 Location: near Detroit, Michigan, USA

Posted: Mon Feb 20, 2006 3:04 am Post subject: More puzzles 


Steve, et al:
There are two recent David Bodycombe puzzles posted in the "4 Star Constellation" thread.
Keith 

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Marty R.
Joined: 12 Feb 2006 Posts: 5499 Location: Rochester, NY, USA

Posted: Wed Feb 22, 2006 12:42 am Post subject: 


Quote:  Anyway, the current puzzle is interesting. What I found is that there are two unique rectangles that eliminate some possibilities, but they are not needed. Even if you find them, you still need to find an XYwing in the top right area of the puzzle. Then, like last week's puzzle, the final step is either a BUG (which assumes uniqueness), or (another) XYwing (which does not). 
I solved this one without any trouble, using nothing more than hidden singles, naked pairs and triples and locked candidates.
It mystifies me how I solve some of these puzzles that I think I shouldn't and get stuck on others that I think I ought to solve.
There seems to be no middle ground; I either solve via the techniques mentioned above or else I get stuck. I seem to only very rarely find opportunities to use all these other techniques which I've read about and studied. 

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