View previous topic :: View next topic 
Author 
Message 
Marty R.
Joined: 12 Feb 2006 Posts: 5470 Location: Rochester, NY, USA

Posted: Wed May 24, 2006 1:41 am Post subject: Rectangle question 


The grid below just shows rows 6789. Note the 39 rectangle in c78, with the 39s on a diagonal. I know for sure that r6c7 must be 6 or 7 or that r9c8 must be 2.
But can I do more? I was ready to do all sorts of candidatezapping on the assumption that the 39 diagonals were conjugate. But thinking further, I can't see a reason why they couldn't be both 3s or both 9s.
At this point, I can't see anything to do with it other than starting a tripleimplication chain (is that the right term?) on the possibilities at the end of the first paragraph above, if I were so inclined.
Code:  1257 237 8 1257 1267 4 3679 39 26 

6 12 259 124 1234 1235 8 29 7 
3 1278 27 1278 9 1278 46 5 46 
25789 278 4 2578 2378 6 39 239 1 
 


Back to top 


Myth Jellies Guest

Posted: Thu May 25, 2006 9:04 am Post subject: 


From just what you have here, the strong links between the 9's in row 6 and the 3's in row 9, along with the 29 cell in r7c8 allows you to eliminate the 3 in r6c7, and the 9 in r9c8.
Code:  1257 237 8 1257 1267 4 A3679 B39 26 

6 12 259 124 1234 1235  8 E29 7 
3 1278 27 1278 9 1278  46 5 46 
25789 278 4 2578 2378 6 C39 D239 1 
 
D2 = A(6or7)  A9 = B9  E9 = E2  D2 loop implies...
A (r6c7) is either 6,7, or 9;
Either B or E is 9, so no other cell in c8 (including r9c8) can be 9;
Either E or D is 2, but that should be handled already by the triples in BDE and CDE. 

Back to top 


Guest

Posted: Thu May 25, 2006 9:16 am Post subject: 


Oops, there is no strong link in the 3's. But the reductions still hold since I did not use a strong link between 3's in the chain/loop. 

Back to top 


Myth Jellies Guest

Posted: Thu May 25, 2006 9:25 am Post subject: 


The 3's are locked in r9b9, so you can eliminate the 3 in r9c5. The two aforementioned strong links with 3 and 9 are all you need to make the eliminations.
Code:  1257 237 8 1257 1267 4 A3679 B39 26 

6 12 259 124 1234 1235  8 E29 7 
3 1278 27 1278 9 1278  46 5 46 
25789 278 4 2578 278 6 C39 D239 1 
 
D2 = A(6or7)  A9 = B9 chain kills the 9 in cell D
A(6or7) = D2  D3 = C3 chain kills the 3 in cell A 

Back to top 


Marty R.
Joined: 12 Feb 2006 Posts: 5470 Location: Rochester, NY, USA

Posted: Thu May 25, 2006 5:05 pm Post subject: 


Thank you, Mr. Jellies, for taking the time to help.
Quote:  D2 = A(6or7)  A9 = B9  E9 = E2  D2 loop implies...
A (r6c7) is either 6,7, or 9;
Either B or E is 9, so no other cell in c8 (including r9c8) can be 9;
Either E or D is 2, but that should be handled already by the triples in BDE and CDE. 
Unfortunately, I just don't understand the notation. For example, D2=A(6 or 7): does than mean if D(r9c8)=2, then A(r6c7) must be 6 or 7? If so, I don't see how the 9 was excluded from A.
Then,  A9=B9: obviously, both A and B can't be 9, so there's clearly another meaning. Like I said, I don't understand the notation, don't understand the meaning of the hyphens and don't see a loop, if loop means continuous closed chain.
I don't want to seem like an ingrate, and I want to avail myself of the help, but I'm just unable to do because the notation is unfamiliar to me and I just can't figure it out on my own. Sorry. 

Back to top 


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

Posted: Thu May 25, 2006 7:33 pm Post subject: I'll take a stab at it. 


Hi, Marty!
I wasn't going to reply to this one, because I'd rather look at the entire grid, and not just a (possibly unhelpful) piece of it. But since MJ pointed the way, I'll take a stab at it.
Here's the fragment you posted.
Code:  1257 237 8 1257 1267 4 3679 39 26 

6 12 259 124 1234 1235 8 29 7 
3 1278 27 1278 9 1278 46 5 46 
25789 278 4 2578 2378 6 39 239 1 
 
 The cell r7c8 must contain either a "2" or a "9". Let's look at the grid both ways.
If r7c8 = 2 the grid looks like this.
Code:  1257 237 8 1257 1267 4 3679 39 26 

6 1 59 14 134 135 8 2 7 
3 1278 27 1278 9 1278 46 5 46 
25789 278 4 2578 2378 6 39 39 1 
 
Now we have the "UR" pattern almost fully resolved, and we can conclude that {3, 9} can be excluded from r6c7, to avoid the "deadly pattern." That leads to this position (because "9" is now unique in row 6):
Code:  1257 3 8 1257 1267 4 67 9 26 

6 1 59 14 134 135 8 2 7 
3 1278 27 1278 9 1278 46 5 46 
25789 278 4 2578 2378 6 9 3 1 
 
OK, that's the first case. What does it look like if r7c8 = 9? Here we can make the reductions right away.
Code:  1257 237 8 1257 1267 4 9 3 26 

6 12 25 124 1234 1235 8 9 7 
3 1278 27 1278 9 1278 46 5 46 
25789 278 4 2578 2378 6 3 2 1 
 
If we compare these two resulting positions we see what MJ was explaining:
 There cannot be a "3" at r6c7.
 There cannot be a "9" at r9c8.
I hope that's clearer, Marty. dcb 

Back to top 


keith
Joined: 19 Sep 2005 Posts: 3257 Location: near Detroit, Michigan, USA

Posted: Thu May 25, 2006 8:33 pm Post subject: Please post 


Marty,
Can you please post the original starting point, and the complete grid? I'd like to examine it.
This is very much like a rectangle type that has been discussed in the Solver's Forum over the past few weeks. The pattern consists of the rectangle, plus a cell that contains one of the "deadly" values plus one of the "extra" values from the rectangle. Then, look for a short chain.
Keith 

Back to top 


Marty R.
Joined: 12 Feb 2006 Posts: 5470 Location: Rochester, NY, USA

Posted: Thu May 25, 2006 8:43 pm Post subject: 


Quote:  I wasn't going to reply to this one, because I'd rather look at the entire grid, and not just a (possibly unhelpful) piece of it. 
David, I certainly would have posted an entire grid if I was asking for general help from my stuck position. My initial goal was an answer regarding UR theory and I thought what I posted was sufficient for that.
Quote:  I hope that's clearer, Marty. 
It is. I'm almost positive I tried that little DIC. However, on the blank grids I use to test chains, there were no cells that were the same for both values in the base cell, thus I thought the DIC was of no help. The fact that it showed the absence of a couple of numbers was too subtle for me, but I now know to be on the lookout for such things.
Thank you.
P.S. MJ, if you'd be willing to briefly explain the notation, I'd appreciate it. 

Back to top 


Marty R.
Joined: 12 Feb 2006 Posts: 5470 Location: Rochester, NY, USA

Posted: Thu May 25, 2006 9:44 pm Post subject: 


Keith, your post wasn't on here when I was making mine. I can't recreate the point where I was stuck, unless I start it over again, and I don't want to do that. It involved coloring, two XWings, a hidden pair, a rectangle and a DIC to get to where I was stuck, plus it was my second or third try at it, so I want it to be history.
Update: removing the "3" and "9" from the two cells was all it needed to easily solve the rest of it. Thank you again, MJ and DCB.
If anyone's interested, it was the May 12 "Tough" puzzle from www.sudoku.com.au . That site's puzzles give me trouble; I find them considerably harder than the "Nightmares." 

Back to top 


keith
Joined: 19 Sep 2005 Posts: 3257 Location: near Detroit, Michigan, USA

Posted: Thu May 25, 2006 11:26 pm Post subject: Thanks! 


Marty,
Thank you. I took a look at the original puzzle and, yes, my solver did not recover the position you posted.
The pattern to look for is this: A UR plus a cell involving one deadly value (one of <39>) and one "extra" value (one of <267>). The "plus" cell here is <29>.
Tabulate the possible solutions of the five cells. Note which candidates are "missing" from the list of all solutions. These can be eliminated. Or, look for a chain.
Look at it this way: If R6C7 is <3>, it forces the nonunique rectangle.
If R9C8 is <9>, it forces the opposite nonunique rectangle, because of the strong link on <9> in R6.
Good stuff!
Keith 

Back to top 


Marty R.
Joined: 12 Feb 2006 Posts: 5470 Location: Rochester, NY, USA

Posted: Fri May 26, 2006 12:29 am Post subject: 


Thanks Keith, I'll add that to my burgeoning collection of solving techniques. 

Back to top 


Myth Jellies Guest

Posted: Fri May 26, 2006 2:52 pm Post subject: 


Myth Jellies wrote:  The 3's are locked in r9b9, so you can eliminate the 3 in r9c5. The two aforementioned strong links with 3 and 9 are all you need to make the eliminations.
Code:  1257 237 8 1257 1267 4 A3679 B39 26 

6 12 259 124 1234 1235  8 E29 7 
3 1278 27 1278 9 1278  46 5 46 
25789 278 4 2578 278 6 C39 D239 1 
 
D2 = A(6or7)  A9 = B9 chain kills the 9 in cell D
A(6or7) = D2  D3 = C3 chain kills the 3 in cell A 
Explaining the notation:
A, B, C, ... identify cells or a set of cells.
= represents a strong inference between two candidate premises (D2=A(6or7) means at least one of "D equals 2" and "A equals 6 or 7" is true)
 represents a weak inference between two candidate premises (both premises cannot be true)
Chains which end in strong inferences and alternate between strong and weak imply that there is a strong inference between the endpoints of the chain.
The strong inference between D and A comes about because of the existance of the unique rectangle in ABCD. 

Back to top 


Marty R.
Joined: 12 Feb 2006 Posts: 5470 Location: Rochester, NY, USA

Posted: Fri May 26, 2006 3:52 pm Post subject: 


Thank you MJ, it has become much clearer now. 

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
