I am playing this Slitherlink app on Android, and get stuck on this puzzle at the following state: enter image description here

I made a lot of progress on the right side, but after this I couldn't find anything that I can do, without potentially guessing a very long chain.

Can anyone help me reason the next logical step from this state?

  • $\begingroup$ off-topic, but may I recommend this one? play.google.com/store/apps/… It is completely free and open source, and has a lot of the different puzzles including Slitherlink. Only weirdness about it is that the names are all different, Slitherlink is called "Loopy" for example. $\endgroup$ Apr 29 at 7:41
  • $\begingroup$ @htmlcoderexe: I recommend those puzzles as well! They can be played online at Tatham's website and/or downloaded as stand-alone executables. And my favourite Loopy variant is Cairo. $\endgroup$
    – user21820
    Apr 29 at 17:20

2 Answers 2


There are a few places where you can place lines leaving a "2" square.
E.g in the top left:

One edge marked A must be a line, so exactly one B must also be a line to have two lines around the 2.
So C must be a line for the continuation from B.

Similarly, one D must be a line to continue the loop, so E must be a line to have two lines around the 2.

Then, one F must be a line to continue from E. That's the only line required for the "1", so both G's must be blank.

enter image description here

And a less obvious example of a pattern where lines touch a 2-square on diagonally opposite corners:
Because of the known lines, the edges marked A cannot both be lines, and the edges marked D cannot both be lines.
So exactly one A and one D must be a line to have two lines around the 2-square. Both of those continue known lines, so B and E must both be blank.
(And then C and F must both be lines to complete the other 2-squares)

enter image description here

And some similar stuff on the right hand side:
One A must be a line for the 1-square, so B must be a line to continue the loop.

And similar to the top-left corner, one of C must be a line for the 1-square, so one of D must be a line to continue the loop.
Then one of E must be a line to have two lines around the 2-square, and F must be a line to continue the loop.
So G must be blank and all the remaining edges of the 3-square must be lines.

enter image description here

  • 1
    $\begingroup$ Ah, interesting, I guess I shall try this technique on other parts of the board too. Thanks! $\endgroup$
    – justhalf
    Apr 28 at 15:28

For the 1 near the bottom left, the left or bottom edge must be included because of the top-right corner of the bottom-left 3. So the top of the 1 is not an edge, which determines the 3 edges for the neighboring 3.

enter image description here

I see now that this method is mentioned here.

  • $\begingroup$ There was a recent question related to this $\endgroup$
    – bobble
    Apr 28 at 17:11
  • $\begingroup$ @bobble it's different, right? The other question is when we know one of the vertices on the 3 box is definitely an endpoint, so the opposite corner must not be an endpoint. For this one, it's about one edge of the 3 box being confirmed blank. $\endgroup$
    – justhalf
    Apr 29 at 4:34
  • $\begingroup$ From your explanation, it looked like you were using the rocket pattern with the corner 3 and the 1 to rule out the 1's top edge as a possibility. $\endgroup$
    – bobble
    Apr 29 at 4:46
  • $\begingroup$ Ah, you mean the reverse, and the corner 3. Hmm, well, I think I can accept that then, although I feel this deduction is much easier than the rocket pattern, where the 3 box is missing one edge. (the fact that this is easy is what made me slightly annoyed that I missed this, actually, haha, I imagine I wouldn't be as annoyed if it were real rocket pattern) $\endgroup$
    – justhalf
    Apr 29 at 6:26
  • $\begingroup$ Yes this can be viewed as the rocket pattern, using the two known corner edges of the corner 3 to deduce the two non-edges of the 1. But I used only the top-right corner based on the third known edge, so it is a bit simpler. $\endgroup$
    – RobPratt
    Apr 29 at 13:22

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