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

Posted: Tue Jan 31, 2006 4:23 pm Post subject: "Nightmare" for January 30, 2006 


Here's a very tough puzzle from the "Nightmare" site.
Code:  . . 6 . 1 . . 4 .
. . . . . . . 7 .
3 1 . . . . 5 . 6
. . . . 2 9 . . 7
. 8 . . 3 . . 1 .
7 . . 5 6 . . . .
5 . 9 . . . . 8 3
. 4 . . . . . . .
. 2 . . 8 . 9 . . 
Just to make it easier, here's what the matrix looks like after we make all the relatively simple eliminations.
Code:  289 579 6 23789 1 23578 238 4 289
2489 59 2458 234689 459 234568 1238 7 1289
3 1 2478 24789 479 2478 5 29 6
146 356 145 18 2 9 468 356 7
269 8 25 47 3 47 26 1 259
7 39 124 5 6 18 248 239 2489
5 67 9 12467 47 12467 1247 8 3
168 4 1378 123679 579 123567 1267 256 125
16 2 137 13467 8 134567 9 56 145 
I'm posting this puzzle because there's a very slick trick that allows one to set the first additional digit. Even with that digit set the puzzle is quite challenging. Oh  this is a new one for me. I've never seen this particular combination (the very slick trick) in a sudoku before. And I didn't spot the new pattern on my own  Ruud van der Werf had to point it out before I could see it. Thanks for the new idea, Ruud! dcb 

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

Posted: Fri Feb 03, 2006 1:57 pm Post subject: It’s a Swordfish with a Fin 


I researched the puzzle. Just as well because, left to my own devices, I would not have found the solution in a year. It is certainly an elegant argument. Thinking about it for a couple of days brought a standard alternative approach to mind. An illustration may help those unfamiliar with fins.
Start with an ordinary swordfish based on three rows. The pattern is:
… A … B … C …
… D … E … F …
… G … H … I …
A, B, .. are the nine cells which may admit an entry X, three in each row.
Now imagine cell G replaced by boxG, the box which contains cell G. Thus the bottom row still has only three places for X and they still line up with the rows above. It is just that one of the “places” is three cells lined up in a box rather than a single cell. BoxG is usually called the fin of the fish. The fin can have places for as many X’s as you like.
Logic:
(1) By construction the top two rows between them must contain two X’s.
(2) If either cell A or cell D contains X, X cannot be placed in the intersection of column AD and boxG.
(3) If neither cell A nor cell B contains X, then either cell B and cell F contain X or cell C and cell E contain X. Whichever obtains, neither cell H nor cell I contains X so X must be on row HI and in boxG.
(4) Thus in all cases X can be excluded from two cells in boxG. These cells comprise the intersection of column AD and boxG excluding cell G itself.
There is such a finny swordfish in David’s matrix. X is 7. The top two base rows are row 1 and row 5. The bottom row is row 9 and the fin is box (3, 1), aka box 7. So 7 can be excluded from r7 c2, which must then contain 6, and the difficult step in the puzzle is resolved.
The “fin” argument is quite general. It applies to any m x m fish (m = 2 being an xwing, m = 3 a swordfish, etc). Also any cell in the fish’s skeleton may be replaced by a box. I do not however think that a useful fish can have two fins.
This standard approach is less elegant than the one David has in mind but it may be more practical. I can just about spot an xwing in good light but solvers practised at identifying fish may find the fin adds little extra difficulty.
I do not know who to credit with the discovery of the fin and should very much like to if anyone can tell me. Possibly the idea has simply been part of the folklore for some time.
Steve 

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someone_somewhere
Joined: 07 Aug 2005 Posts: 275 Location: Munich

Posted: Sun Feb 05, 2006 11:28 am Post subject: 


Hi,
Yup the digit 7 is the key for this Sudoku.
Looking it from the point of view of the DIC (Double Implication Chains) by starting from a pair, it can also be solved.
Here the key for the solution starting from r7c2 = 7
we follow 3 pathes:
(A) r7c2 = 7, r7c5 <> 7
(B) r7c2 = 7, r7c7 <> 7, r8c7 = 7, r8c5 <> 7
(C) r7c2 = 7, r8c3 and r9c3 <> 7, r3c3 = 7, r3c5 <> 7
and from (A), (B) & (C) there is no place for a 7 in column 5 !!!
so we can exclude 7 from r7c2, and the rest is just an exercise.
P.S. I have found also a 4 star constellation to exclude 6 from r4c8, but this is not breaking the puzzle: r5c3 = 2, r5c7 <> 2, r5c7 = 6, r4c8 <> 6
and second path: r5c3 = 5, r5c8 <> 5, r4c8 = 5, r4c8 <> 6.
see u, 

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