A few notes: When we speak of a numbered cell being "filled", we mean that it
has all of the neighbors in the path that it can have. Similarly, when we say that a
numbered cell "can be filled", we mean that the remaining unmarked neighbors
must all be on the path in order for it to be filled. Finally, when we say that a
cell would be "broken", we mean that we have proven that the cell would need
more neighbors than it is allowed.
Also, for shorthand, we will specify that Rule X is that every cell in the path has two
neighbors that are adjacent to it in the path, and Rule Y is that if a cell has only two
neighbors that are neighbors of each other, then the original cell cannot be on the path.
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If A6 were not in the path, then A5 and B7 wouldn't be [Y], but that
breaks B6. Therefore, A6, A5, and B7 are in the path. If C6 were in the path, then B5 or
C7 would have to be [X], but that breaks B6, so C6 isn't on the path. If A1 were in the
path, then A2 and B1 would be too, but this breaks B2, so A1 is not in the path.
Therefore, A2 and B1 aren't either [Y].
If A4 weren't in the path, then A3 and B3 wouldn't either [X,Y], but that breaks B4.
Therefore, A4, A3, and B3 are in the path. This fills B2. C4 is in the path [X], so B4 is
also filled. We can fill B6 now.
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If D6 were in the path, then D5 and E7 would be too [X], but this would
break E6, so D6 is not in the path. We can fill D7 now. F7 must be in the path [X,E7], so
E6 is filled.
F8 cannot be in the path [X,Y], so F9 can be filled. C8 isn't in the path [Y]. The only
way to rectify all of the paths in this region is if the path B7, C7, D8 ... continues
through D9, E10, F11..., and the path F7, E7, E8... continues through E9, F10.... These
paths cannot connect in this area without violating Rule Y, so the first path continues
F11, G11, H11, I11... and the second path continues F10, G10.... J10, of course, can be
filled in. J11 must connect to I11 and K11, and I10 to H10 and I9.
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If J5 or K7 were on the path, then J6 would have to be as well [X,Y], but
this would break K6. Therefore, only J6 is on the path, which must connect to I5 and J7.
If K8 were on the path, then K9 would be as well, but there would be no way to connect all
of the points in that region without sharp turns, so neither K8 nor K9 is on the path. We
can fill J8.
If H9 were on the path, then G8 would be too, but that would break G9. Therefore, H9 isn't
on the path, so we can fill G9. G8 connects to F7 and H8. If I9 connected to H8, then we
would have an incomplete circuit, so I9 connects to J9 and I8 connects to H8 and I7. C1,
G7, H7, and I6 aren't on the path [Y], and D4 is.
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H) We can fill in E2. If F1 were on the path, then it would connect to E1
and G2, but then F2 would violate [Y], so F1 is not on the path. If G3 weren't in the
path, then G4 could be filled, H3 and G2 would be in the path, but then F3 would violate
[Y], so G3 is on the path. F2 cannot connect to F3 or G3, so it must connect to G2, then
to H3, I4, and I5. F3 connects to G3, H4, and H5.
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If D3 were in the path, then it could only connect to D2 and D4, which
would form an incomplete loop, so D3 is not in the path. E3 can only connect to F3. If F4
were in the path, then G5 would too, but that would break G4. Therefore, G5 is not in the
path, so we can fill G4. The final spots in the path must be E4 and F5. |
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The diagonal path is 11001010110 |
Comments: Solving this one was my magnum opus for the contest. It probably took
me a half hour, so it wasn't worth it from an efficiency point of view, but I definitely
felt like I had achieved something at the end of it. I went down a few incorrect
paths and wound up erasing everything and starting over, so if I had it to do again I
think I would have tried solving half of it in pen and the other half in pencil.