Toroidal Killer

For fans of Killer Sudoku, Samurai Sudoku and other variants

Toroidal Killer

Postby Ruud » Sun May 21, 2006 10:40 pm

Here is a new type of killer: Toroidal.

Siobhan Fahey wrote:This sort of twisted, baby-doll innocence is the underbelly of any beautiful girl you see with makeup.


The nonets wrap around the puzzle. At the far sides, there are some nonets split in 3 parts. C'est la vie.

Image

I would like to give you some tips on how to solve it, but I can't. This puzzle has a unique solution, but it does not yield to the normal suite of killer tactics.

Anyone who solves it is invited to share his or her wisdom with us.

Ruud.
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Postby Smythe Dakota » Mon May 22, 2006 10:24 am

The wrap-around amounts to playing on the surface of a torus.

Have you also considered a Klein bottle (south edge is glued to north edge backwards, east and west edges are glued normally), projective plane (south glued to north backwards, east and west also glued backwards), or Moebius strip (south glued to north backwards, east and west not glued at all)?

Bill Smythe
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Postby motris » Mon May 22, 2006 3:12 pm

I started a little work on this fascinating(ly difficult) puzzle last night. There are always 2 goals with such a venture - 1.) Get to a solution; 2.) Get to a solution with every logical step being completely satisfying. I'm pretty confident I can get to 1 now but goal 2 may be trickier.

After lots of staring, I think the best "weakness" I saw to get in and place about 10-15 digits if you accept a trifurcation is in the lower right. The box R78C78 has to add up to 9. This means that a digit is repeated in this square and, specifically, that R8C7 = R7C8 = X = (1, 2, or 3). If you then consider column 9, you'll find that R6C9 must also be X based on the toroidal shape and the continuation of the cages that already contain an X. So, you have a good starting nucleus of three identical but unknown digits to work with. Also, you'll find that several other digits in the 6-cage and 13-cage and in column 9 are constrained to just 1 distribution for each of these 3 cases which is nice.

Now, looking more at the fixed digit, if X is 1 or 2, you can further go on to prove that all occurrences of X must lie along an intact diagonal, mainly because of the constraints imposed by the sum of the middle column R456C5 being 21 (so X cannot be there) and the 27 cage in the lower left (so X cannot be there) along with some Nishio-like tests. So, if X is 1 or 2, then X is in R9C6, R1C5, R2C4, R3C3, R4C2, R5C1 as well. If X is 3, you cannot as immediately apply this rule as the 27 cage is no longer disallowed for X (the 21 column still is) but you can show that the 3's are either along this diagonal or include R9C2 and R5C3 and 4 other squares you cannot further specify. The proof of all this is a little difficult to write out without a picture, but if you get comfortable with the toroidal shapes, I think its possible to see if you try.

So, now I just need to either further constrain X or accept the trifurcation and use the 12-15 digits it lets me place to move forward. I didn't see any quick way for the former but would still like to succeed there. I started with the harder case when X = 3, and have made enough progress (15 placements) to say it does not use the diagonal case above. I'm convinced once I get back to it, I will either find a solution with X=3, or by eliminating that case, can use the long diagonal and X=1 or 2 to make enough progress to solve the puzzle as once I have a couple digits, I can start to use a bunch of other rules for placing digits like some of the toroidal tricks I've come up with before.

Of course, all of this work was done late last night before I went to bed, and it will be a little bit of time before I can work at it again, but I think this is a reasonable way to think of starting. I spent a bunch of time before I saw these three forced linked digits just reducing candidates in odd ways and think another potential route is to constrain the 1's and 9's as many cells cannot hold one of the two extremes and then to further remove candidates by coloring/Nishio like routes.

Anyway, my two cents. Maybe some puzzles are made to not be solved, but I'm not sure I cannot finish this one.

Thomas Snyder
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Postby Pyrrhon » Mon May 22, 2006 6:12 pm

I know it is a little bit complicated with cross-posting. But there is an solution now in another forum.

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Postby motris » Mon May 22, 2006 6:49 pm

I wasn't seeing it last night, but the linked solution focuses on the same place I've been looking so it definitely seems the best way in. I had the 3 pairs [231] [31(45)], [312] [14(35)], and [123][24(16)] for the 6/13-cages - what I was calling my trifurcation point - but, because I started my elimination process on the correct route (with X = 3), I hadn't run into the simple eliminations for when X = 1 or X = 2. Should have started in numerical order:) .
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Postby motris » Tue May 23, 2006 2:40 pm

Finished it quickly from where I left off last night. It ended up having one repeated left/right pair of digits as I've discussed previously for the toroidal type, but there was never a reason to really use this property as too many digits were placed before it was clear.

Anyway, thanks Ruud for this excellent challenge.

Thomas Snyder
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Postby Ruud » Tue May 23, 2006 6:10 pm

Thanks for the feedback.

The next challenge may include wraparound cages on a projective plane. Gives you solving strategies like: There are 2 outies for Nonet 1 in R8C9 & R9C8...

Ruud.
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