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I'm stuck on this game of Slitherlink. What deduction (or deductions) can be made from here? Am I missing something obvious?

I verified that everything I have so far is correct by looking at the solution.

This is game 6x6t4:a4143b22a3a2g4d32a3a3b0a1b34b2h32a21b2a (width 6, height 6, grid type "Cairo", difficulty hard) in Simon Tatham's "Loopy", version 20190415.e2135d5.

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Take a look at the bottom middle section:


enter image description here

I claim that edge H cannot be used.

F and G must be used together. So H would have to go with either I or J. But if J is used, then E must be used, and so EJK would complete a tiny loop. And if we use HI, then we also must use AB, and now the vertex between D and E can't go anywhere.

So great, H can't be used. What does that tell us?

Well, now either IJ or FG must be used: exactly one of those two. So take a look at the vertex near point EFJ. The loop enters from either F or J, and does not exit from the other, so it must use edge E.

And now more progress can be made using various clues in that area and on the left. This should be enough to get you unstuck.

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    $\begingroup$ Perhaps more directly, using DC leads to not using A or F, contrary to the clue there. $\endgroup$ – Daniel Mathias Apr 15 at 17:23
  • $\begingroup$ Good one, I would never have found that. But I found another way (using a different technique) while trying your solution, so I added another answer. $\endgroup$ – Martin Herrmann Apr 15 at 20:11
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    $\begingroup$ Yet another way, a line at FG quickly leads to a contradiction, and the rest of the area can be solved from there. $\endgroup$ – Kruga Apr 16 at 8:06
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Inspired by Deusovi's answer, I found another deduction in the same general region - one that I, personally, find simpler.

enter image description here

First, look at edges

A and B

We can deduce that

they can't both be used because that would inhibit CD and not leave enough options for the 2. They also can't both be unused because then CDE would either all have to be used (too many for the 2) or all have to be unused (not enough left for the 2). So we know that of A and B, exactly one must be used.

Now look at the highlighted path. We know that any closed path

must cross an even number of used edges, because each crossing takes us from inside the loop to the outside or vice versa.

We know that

Exactly one of A and B must be used (as shown above) and exactly one of F and G must be used (as per the clue).

It follows that

H must also be used to make the number of edges crossed by the path even.

The rest is straightforward.

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  • $\begingroup$ What do you mean by "cross an even number of used edges"? $\endgroup$ – Matsemann Apr 16 at 8:40
  • $\begingroup$ In the image I posted, the red path crosses edges A, B, F, G, H, and two unlabeled edges between H and A. Of the unlabeled edges, one (black) is marked as used an one (gray) is marked as not used. All of the labeled edges are unknown (yellow). Of these edges (A, B, F, G, H, plus 2 unlabeled), an even number must be used (black) in the final solution. $\endgroup$ – Martin Herrmann Apr 16 at 17:23
  • $\begingroup$ I see. Interesting, will have to keep that trick in mind for later. Thanks for explaining. $\endgroup$ – Matsemann Apr 16 at 17:49

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