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This puzzle is part of the Monthly Topic Challenge #13: Variety Slitherlinks.


Have you ever wondered why there has to be an "inside" and an "outside" to every slitherlink loop?

Well, with a bit of rule-bending (grid-bending really) everything can be on the inside (or the outside if you prefer). All you have to do is wrap your grid around a torus so that it has no edges.

And just to add a bit more variety: Now that we have every cell on the same side of the loop, let's also require that there be a unique shortest path (travelling orthogonally) between every pair of cells. the path through the cells is narrow and twisty - no big open spaces.*

  • Draw a single, non-intersecting loop that only consists of horizontal and vertical segments between the dots.
  • The edges of the grid are joined (left-to-right and top-to-bottom) so that the grid is topologically a torus. The loop can travel off the left/right side of the grid and back on the right/left side (and similarly top/bottom).
  • Numbers inside a cell indicate how many of the edges of that cell are part of the loop.
  • There must be a unique shortest path between every pair of cells (travelling orthogonally and not crossing the loop).
  • The path must be "narrow" - there must be no contiguous 2x2 block of cells inside the loop.*

Grid for Mobius slitherlink on a torus

Solve on Penpa here To get the solution checker to work, the left/right and top/bottom joined edges must be consistent (draw the loop on both where it travels along the edge of the grid).

* My attempt to hide the "no 2x2" restriction behind a unique path constraint failed because there is a cycle in the path that goes all the way around the torus. Trying to fix this with "shortest" also fails for even-width grids like this one because the total cycle length is even and for suitable pairs of points, the shortest distance is the same either way round.

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  • $\begingroup$ Oh, sorry! I'll repost in rot13: rot13(Fbzrguvat V'ir abgvprq: Sbe rirel 2k2 frpgvba bs gur tevq, gur fyvgureyvax unf gb cnff guebhtu gung, bgurejvfr gurer jba'g or n havdhr cngu.) $\endgroup$ Aug 15, 2023 at 15:28
  • $\begingroup$ I'm confused about the torus requirement. Do the top and bottom edges have to have the same configuration? $\endgroup$ Aug 15, 2023 at 15:53
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    $\begingroup$ And, sorry to nitpick, but a "Möbius strip" has a twist. To make a Möbius strip you should for instance match the top from left to right with the bottom from right to left, and leave the sides disconnected. If you mean the path has a single side, it is not the case. If the path were a wall and you painted one side of it from end to end, so to speak, you would still leave one side unpainted. This being said, it is still a great puzzle. $\endgroup$
    – Florian F
    Aug 15, 2023 at 19:44
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    $\begingroup$ @FlorianF nitpicking is fine. I was just using a bit of poetic license for having a closed loop that doesn't have an inside and outside any more. But you are correct that the loop itself still has two "sides". $\endgroup$
    – fljx
    Aug 15, 2023 at 20:17
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    $\begingroup$ @FlorianF hmm. I may have to give up on path uniqueness, and just openly state the required restriction. (Which is pretty obvious anyway.) $\endgroup$
    – fljx
    Aug 15, 2023 at 21:16

1 Answer 1

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What a great puzzle! The following is a solution:

solution

The key:

is to look at the 1s. In most cases they are not very important, but here 1s, alongside the shortest path rule (which turns into a restriction that every 2x2 square has to have a path through it) provide a large amount of information. Specifically, if two 1s are diagonal, the line cannot be 'around' them, or else it would lead to a contradiction.

We start by using the above rule and the

zero in the bottom-right corner. We can then easily deduce that there have to be lines pointing in to the zero. Similarly the setup of 1s in the left side of the grid allows for us to deduce a large amount of information.

I have liberally used this rule:

To keep the shortest-paths/2x2 grid law intact, if in a 2x2 square two X's are drawn in the following formation, then we must have two lines pointing out. rule

I started with

start point and made deductions to get to the final solution.

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  • $\begingroup$ Well done. This is the intended solution. $\endgroup$
    – fljx
    Aug 16, 2023 at 8:22

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