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Black Belt Sudoku - Frank Longo - another killer

 
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boweasel



Joined: 15 May 2006
Posts: 5

PostPosted: Sun Jun 11, 2006 5:11 pm    Post subject: Black Belt Sudoku - Frank Longo - another killer Reply with quote

Okay, here's another one I seem unable to solve by what I consider reasonable methods. I can find no usable unique rectangles, X-wings, Y-wings, XY-wings, or swordfish.

I HAVE solved the puzzle. However I've only been able to do so by using that most heinous of techniques - trial and error. If I look at r1c5, only one of the three candidates leads (eventually) to a solution.

But I hate this technique to the point that I'm seriously considering chucking the book. I have no problem with difficult puzzles that use solving methods unknown to me, and you folks have been great at helping me so far.

Is there a method for this puzzle, and if so, and if it involves coloring how do I recognize where to begin?
Code:

37 12  5    8  279 29   6  4   13
37 9   4    27 6   1    8  235 35
6  128 18   4  3   5    27 29  179

1  5   6    9  27  3    27 8   4
89 4   89   27 1   6    35 235 357
2  3   7    5  8   4    9  1   6

59 7   39   1  4   8    35 6   2
58 18  138  6  259 29   4  7   359
4  6   2    3  59  7    1  59  8
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keith



Joined: 19 Sep 2005
Posts: 3136
Location: near Detroit, Michigan, USA

PostPosted: Sun Jun 11, 2006 5:54 pm    Post subject: Reply with quote

In the position you posted, there is an X-wing on <2> in R25. You can eliminate <2> from R3C8, which solves the puzzle.

Keith
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Marty R.



Joined: 12 Feb 2006
Posts: 5065
Location: Rochester, NY, USA

PostPosted: Mon Jun 12, 2006 3:23 pm    Post subject: Reply with quote

Quote:
Is there a method for this puzzle, and if so, and if it involves coloring how do I recognize where to begin?


It can also be solved by coloring, but Keith's X-Wing is probably more straightforward. I just learned coloring a couple of months ago, so I'm hardly an expert.

Where to begin? I don't know, other than to scan, looking for rows/columns/boxes with just two possibilities for a number. In this case, start an ab chain of 2s beginning at r3c7.

r3c7=a-->r4c7=b-->r4c5=a-->r5c4=b-->r2c4=a-->r2c8=b.

The ab in r3c7 and r2c8 force the elimination of "2" from r3c8, which is the same elimination as the X-Wing, so it, of course, also solves the puzzle.

I offer this up just as an exercise in coloring. When I saw your grid I stubbornly tried to find an alternate solution and this coloring chain took some time to see, it didn't exactly jump off the page at me.

P.S. As far as where to begin, this one's something I always look for: starting from a group with three possibilities, hoping that by both starting and ending there the third can be eliminated.
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boweasel



Joined: 15 May 2006
Posts: 5

PostPosted: Wed Jun 14, 2006 6:53 pm    Post subject: Reply with quote

Keith, Marty:

Thanks for your help. I can't tell you how long I stared at that puzzle, and never, EVER saw that incredibly simple X-Wing.

I'm ashamed to admit it, but I even made up grids for each number, like here is every 1, here is every 2, etc. But by that time I was looking for something more exotic than a simple X-Wing, so the grids didn't help.

I guess I'll keep the book...

Thanks again,
Bo
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