# How to get started on this Slitherlink (No Swiss)?

This is a puzzle from Round 4 of WPF's Puzzle Grand Prix this year.

It's a Slitherlink with the additional rule that "Every Swiss cross-shaped arrangement of five cells must contain at least one cell inside the loop and one cell outside the loop." What I find particularly interesting (i.e. frustrating) about it is how I can't make any progress at all (besides the obvious 2 in the lower left)! I've realized that

the no-Swiss cross rule implies that every cell not on the edge needs at least one of its edges filled in,

but even this has gotten me no further progress. For what it's worth, I solved the previous puzzle (admittedly worth almost a quarter of the points) without using any "special" tactics, i.e. just as a normal Slitherlink with an additional restriction.

I would appreciate some tips on how to get started, as well as some thoughts on how I might have derived them myself.

• Is this an ongoing competition? Jul 10 at 2:40
• No, this competition ended in April. (The Puzzle GP is about to start Round 7 for this year.) (Also, the solutions are out.)
– fish
Jul 10 at 2:41

That is a very satisfying slitherlink puzzle to solve.

The way I started is to look at a 3 with two adjacent 1s.

Suppose we cross out two of the edges of one of the 1s as shown here:

This means that the edge for that 1 is actually one of those crossed out sides, so I should have eliminated the other two sides instead. This goes for the other 1s in the grid as well, leading to this:

Although a little progress can be made at the top right corner, the main breakthrough for me came from the chains of 2s at the top left. The two 2s marked here must have a path going diagonally top-left to bottom-right (i.e. either top and right edges or left and bottom edges filled in) because the 1 blocks the other diagonal.

The chain through the 2 on the top row forces the top edge of the third box on the top row to be chosen:

This then allows you to fill in a lot of the top left quarter of the grid, and eventually the whole top half of the grid. It is only when you then start on the sparser bottom half that the Swiss cross rule comes into play.

• I hadn't known of this 1-3-1 L pattern, thank you for sharing!
– fish
Jul 10 at 15:33

Another way without trial and error:

Using these techniques, we can finish the puzzle:

1. Paint the dots leading to a filled line. Note that two lines are around them; one leads to one and one starts from it. If you don't know whether it's connected or not, note that there must be an even number of lines connected to it.

2. Mark the lines as a, b, c etc. depending on the possibility of them being filled. If multiple lines are guaranteed to be filled or not at the same time, give them the same name. If given two lines are adjacent and only one can be filled, mark the corner between them as 1.

3. If there are two equivalent lines (with the same mark) and an even number of filled lines around a square or leading to a dot, the remaining two must also be equivalent to each other. If it's an odd number, the remaining two must be the opposites. If there are only two options leading to a dot (the rest are eliminated), they're also equivalent to each other. Paint new dots accordingly.

4. The no-Swiss rule. If some of the options around a square are eliminated, do 1 and 2 to the remaining lines accordingly with 3 in the mix.

Here's the picture (the green lines are after taking 4 into account):