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

Below is a grid containing multiple small slitherlinks. These can be independently solved to form clues for the large metaslitherlink.

The following standard rules apply to all slitherlinks:

  • Connect adjacent dots with vertical or horizontal lines to make a single loop.
  • The numbers indicate how many lines surround it, while empty cells may be surrounded by any number of lines.
  • The loop never crosses itself and never branches off.

The solved metaslitherlink should reveal a number, which is the final answer to this puzzle.


  • 2
    $\begingroup$ Nice twist on the small slitherlinks connection to the big one! $\endgroup$
    – justhalf
    Commented Aug 2, 2023 at 2:55

1 Answer 1




Let's first solve the small slitherlinks:

small slitherlinks

For most of the small slitherlinks I think I know what they mean:

small slitherlinks solution

I just don't know what the Pi symbol means. It's not 3.14 rounded to 3, because then there is no solution possible. Maybe a zero, (because you can't do full circles in slitherlink, the Pi symbol is almost a closed circle), but that generates some negative values (for example, R2C3 will be -3. I tried to interpret those as a 5, because then all the calculations made sense, but that gives not a unique final solution. And what is the symbol in R3C4 supposed to mean?
Edit: as pointed out in the comments: it's not a Pi symbol, it is a N for a number not yet known. And N equals to 4. The symbol in R4C6 reads nil. Thanks @Arturo and @Samm82.

If I interpret it as 4, I get the following:

And I think this could be the final answer:

4^4 = 256

  • 2
    $\begingroup$ I got stuck with the N too, but didn't like case-bashing so I left the meta grid unsolved. I think R3C4 is supposed to say "nil" rather that "n-n". $\endgroup$ Commented Aug 1, 2023 at 19:10
  • 7
    $\begingroup$ The "case-bashing" here doesn't seem that bad to me- n has to be at least 3 (or else R2C3 is negative) and is at most 5 (or else R5C5 is negative), so there is only 3 cases to consider (n = {3, 4, 5}); I would try solving the metagrid without the n-based clues and see if any more constraints on them come up. $\endgroup$
    – samm82
    Commented Aug 1, 2023 at 19:28
  • 2
    $\begingroup$ Nice solve, well done! I settled on using the variable 'n' (running the risk that it could be interpreted as 'pi') because other letters, like 'x' are too awkward to draw in a rectangular grid :) $\endgroup$ Commented Aug 2, 2023 at 7:21

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