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Marty R.
Joined: 12 Feb 2006 Posts: 5485 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 


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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 

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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 

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Marty R.
Joined: 12 Feb 2006 Posts: 5485 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. 

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Marty R.
Joined: 12 Feb 2006 Posts: 5485 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. 

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Marty R.
Joined: 12 Feb 2006 Posts: 5485 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. 

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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 

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Marty R.
Joined: 12 Feb 2006 Posts: 5485 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. 

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