First, here’s an updated version of my previous toroidal slitherlink:
That should now be unique for inside/outside solutions. See below for a description of what inside/outside means.
As I mentioned before, Thomas Snyder pointed out that there are two different types of solutions to a toroidal slitherlink: either there is an “inside” and an “outside,” or there is only one “side.”
First, explanation of the term toroidal. It basically means donut-shaped, and that’s what this puzzle is. If you consider a torus like the one pictured at http://en.wikipedia.org/wiki/File:Torus_cycles.png, the purple line represents the top/bottom of the puzzle, and the red line represents the right/left of the puzzle.
It’s a loop, with an obvious inside and outside. You can confirm this by picking a point inside the loop and another outside the loop. The two cannot be connected by a line without the line crossing the loop.
But other loops are possible in a toroidal slitherlink: a vertical line is a loop, for example. In the illustration of a torus, it corresponds to any loop parallel to the red line. Similarly, a horizontal line is also a loop, corresponding to any loop parallel to the purple line. For both a vertical or horizontal line, there is no inside or outside. Any point on the torus can be connected to any other point by a line without crossing the red or purple lines.
Notice how, because the loop is closed, it must intersect the edge twice. If the loop pokes out in two places, it must intersect the edge four times, as shown in the next illustration.
This pattern holds even when the loop intersects the edge at the corner (in this case resulting in six intersections:
Because the loop can never be broken, it must always intersect the edge in pairs. So:
An inside/outside solution will always intersect the edges of the puzzle an even number of times: 0, 2, 4, etc. A non-inside/outside solution will intersect the edges of the puzzle an odd number of times: 1, 3, 5, etc. Be sure only to count each intersection one time: going out the top/in the bottom counts for one, not two. An inside/outside solution contributes to the solution of the puzzle in several ways:
- As stated, the solution must intersect the edge an even number of times. If a partial solution intersects the edge an odd number of times and cannot intersect it any more, it must be wrong.
- Marking squares as inside/outside, all inside areas must be able to connect; likewise all outside areas. So if there is only one way for a square to be colored that allows this, it is determined. This is a large part of the solution of this puzzle.