View previous topic :: View next topic 
Author 
Message 
Marty R.
Joined: 12 Feb 2006 Posts: 5555 Location: Rochester, NY, USA

Posted: Wed Feb 22, 2006 10:09 pm Post subject: At an impasse (again) 


This was the "Tough" puzzle from sudoku.com.au for February 13.
As published:
Code:  63
549
72

658
32
186

28
975
16
 
And here's where I'm stuck. I've used two forcing chains, but they obviously didn't open things up. If someone can give me a hint on the next step...
Code:  49 1 6 3 279 259 457 8 457
38 5 2 4 67 68 137 9 137
7 38 49 59 1 589 345 6 2
6 247 479 127 5 2349 8 1347 179
4589 478 3 167 469 469 2 147 1579
459 247 1 27 8 2349 579 347 6
2 6 47 59 3 459 179 17 8
1 9 8 26 26 7 34 5 34
34 347 5 8 49 1 6 2 79 


Back to top 


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

Posted: Thu Feb 23, 2006 12:27 am Post subject: Here's an idea ... 


Hi, Marty! This is an interesting puzzle.
Code:  49 1 6 3 279 259 457 8 457
38 5 2 4 67 68 137 9 137
7 38 49 59 1 589 345 6 2
6 247 47+9 127 5 2349 8 1347 179
4589 478 3 167 469 469 2 147 1579
459 247 1 27 8 2349 579 347 6
2 6 47 59 3 459 179 17 8
1 9 8 26 26 7 34 5 34
34 347+ 5 8 49 1 6 2 79 
Observe that there are only two ways to fit a "7" in row 9 and in column 3, and that there are also only two ways to fit a "7" in the bottom left 3x3 box (see the "+" and "" signs in the grid above). Clearly we either have r4c3 = 7, or else r9c9 = 7. Either way there cannot be a "7" at r4c9.
Now, with the possibilities at r4c9 reduced to {1, 9} we can see an XYWing pattern in r9c9. r7c8, and r4c9. If r9c9 = 7 then r7c8 = 1. And if r9c9 = 9 then r4c9 = 1. So there can't be a "1" at either r4c8 or r5c8, leaving r7c8 as the only possible spot to place a "1" in column 8.
That's not enough to solve the puzzle, but it does make a bit of progress. dcb 

Back to top 


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

Posted: Thu Feb 23, 2006 1:21 am Post subject: ... and another 


A slightly different approach to the next two steps is:
(a) eliminate 5 from r2c1 using the finned XWing in rows 1 and 5 (which leaves possibles as (79)) and
(b) use the XYWing pivoted on r2c1 with pincers in cells r6c1 and r2c8 to eliminate 4 from r6c8.
There is then only one place for 4 in row 6 and the rest is routine.
Steve 

Back to top 


Marty R.
Joined: 12 Feb 2006 Posts: 5555 Location: Rochester, NY, USA

Posted: Thu Feb 23, 2006 7:27 pm Post subject: Re: Here's an idea ... 


David Bryant wrote:  Hi, Marty! This is an interesting puzzle.
Code:  49 1 6 3 279 259 457 8 457
38 5 2 4 67 68 137 9 137
7 38 49 59 1 589 345 6 2
6 247 47+9 127 5 2349 8 1347 179
4589 478 3 167 469 469 2 147 1579
459 247 1 27 8 2349 579 347 6
2 6 47 59 3 459 179 17 8
1 9 8 26 26 7 34 5 34
34 347+ 5 8 49 1 6 2 79 
Observe that there are only two ways to fit a "7" in row 9 and in column 3, and that there are also only two ways to fit a "7" in the bottom left 3x3 box (see the "+" and "" signs in the grid above). Clearly we either have r4c3 = 7, or else r9c9 = 7. Either way there cannot be a "7" at r4c9.
Now, with the possibilities at r4c9 reduced to {1, 9} we can see an XYWing pattern in r9c9. r7c8, and r4c9. If r9c9 = 7 then r7c8 = 1. And if r9c9 = 9 then r4c9 = 1. So there can't be a "1" at either r4c8 or r5c8, leaving r7c8 as the only possible spot to place a "1" in column 8.
That's not enough to solve the puzzle, but it does make a bit of progress. dcb 
David, I follow your logic in the first paragraph which eliminates "7" from r4c9. However, spotting something like that appears to be beyond my powers of observation at this time. I also follow how you placed a "1" in r7c8, although you didn't say specifically, I believe the XYWing eliminates the "1" from r4c8.
I agree, it's not enough to solve the puzzle, but I'm more interested in learning a new technique, if indeed, I learned it. With the elimination of the "1" in column 8 of box 6, the remaining triple in column 8 allows removal of the other candidate "7" in that box, but that hasn't helped so far.
As always, thanks for the help. 

Back to top 


Marty R.
Joined: 12 Feb 2006 Posts: 5555 Location: Rochester, NY, USA

Posted: Thu Feb 23, 2006 7:33 pm Post subject: Re: ... and another 


Steve R wrote:  A slightly different approach to the next two steps is:
(a) eliminate 5 from r2c1 using the finned XWing in rows 1 and 5 (which leaves possibles as (79)) and
(b) use the XYWing pivoted on r2c1 with pincers in cells r6c1 and r2c8 to eliminate 4 from r6c8.
There is then only one place for 4 in row 6 and the rest is routine.
Steve 
Steve, I don't understand. There was no "5" in r2c1 and the XYWing pivoted in r2c1 doesn't mesh with r6c1 and r2c8. ???
Is there a difference between a "finned XWing" and an "XWing"?
Thanks for taking the time to respond. 

Back to top 


Marty R.
Joined: 12 Feb 2006 Posts: 5555 Location: Rochester, NY, USA

Posted: Thu Feb 23, 2006 10:22 pm Post subject: Solved!! 


Three hours after my previous reply, it was solved with a couple of forcing chains. Thanks again for the little push. 

Back to top 


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

Posted: Fri Feb 24, 2006 1:47 am Post subject: The real alternative 


Sorry Marty: of course you didnt understand it.
The story is this. Studying the puzzle and drafting my note took some time. When I came to post it, David had already answered so I scrubbed my effort. Then I thought it might be worth sending anyway, retrieved it, amended it slightly and sent it. The flaw in the procedure was retrieving the wrong note from the recycle bin. What I thought I sent was:
A slightly different approach to the next two steps is:
(a) eliminate 7 from r4c9 using the finned XWing in columns 3 and 8 (which leaves possibles as (19)) and
(b) use the XYWing pivoted on r9c9 with pincers in cells r7c8 and r4c9 to eliminate 1 from r45c8.
There is then only one place for 1 in column 8.
As for fins, an illustration may be of interest. Start with a normal XWing based on columns:
A
C
.
.
B
D
A and B are the only cells which may admit an entry, X, in the lefthand column; similarly for cells C and D in the righthand column.
Now imagine cell C replaced by boxC, the box which contains cell C. Thus the left hand column still has only two places for X and they still line up with the column to the right. It is just that one of the places for X in the right hand column is three cells lined up in a box rather than a single cell.
Logic:
(1) Either cell A or cell B contains X.
(2) If cell A contains X, X cannot be placed in the intersection of row AC and boxC.
(3) If cell B contains X, then D cannot and the X in boxC lies in column CD.
(4) Whichever of (2) and (3) applies, X can be excluded from two cells in boxC. These cells comprise the intersection of row AC and boxC excluding cell C itself.
The pattern is less useful than an XWing but, if you look for XWings at all, you will spot the finned version just as easily. Incidentally, any m x m fish can have a fin (where m = 2 is an XWing, m = 3 a swordfish, etc). By coincidence a finned swordfish can be applied later in the same puzzle.
Steve 

Back to top 


Marty R.
Joined: 12 Feb 2006 Posts: 5555 Location: Rochester, NY, USA

Posted: Fri Feb 24, 2006 6:29 pm Post subject: 


Thanks Steve, for clearing that up. I've printed out your reply, as I did David's, and will pore over it to see what I can learn. 

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
