by **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