A curious Slitherlink Deduction

Question

In the following Slitherlink, where you need to make one unbroken loop over the grid with the number in each cell matching the number of edges around that cell, I noticed that the 3-2-2-1 2x2 box at the middle top has only one possibility: the 1 having their right edge on as opposed to their bottom edge.

It's a quick deduction if you do trial and error using the bottom edge of the 1, as it quickly leads to contradiction on the top two cells.

But as trial and error method is typically not the first method one would use, I'm curious whether there are another way to deduce that the 1 should have the right edge on, instead of the bottom edge. My usual method of making an inaginary loop and counting edges (must be even) or even the latest trick I learned from this community about inside-outside don't seem to work 🤔

(Or otherwise, how to proceed in from this partially filled grid. I tried trial and error on the bottom left too, but ended up needing to resolve this section as well.)

Answer 1

There are a couple of ways to analyse that area without any guesswork or backtracking, which both reach the same conclusion:

If we colour the cells that we know are inside and outside the loop green and blue respectively.
We know from the lower-left 2 that C must be outside the loop (blue), so there must be a line above it (drawn here in orange).

For anyone wondering how we can work out C is blue:
Exactly one of the remaining unknown edges around the lower-left 2 must be a line.
So one of C and D is on the same side of the loop as the 2, and the other is on the opposite side.
So C and D must be different colours.

Within any contiguous region of the grid there must be an even number of ends.
The region inside the red box currently contains 6 ends.
We know that exactly one of the external connections to this area from the lower-left 2 must be a line (A).
That gives us 7 ends within the red box, so the only remaining unknown edge entering the box (B) must be a line.

Answer 2

Found another way to look at this that is closer to directly answering the first question:

Consider the 2x2 box in question as a whole:
Starting in the bottom left and working anti-clockwise, each pair of edges must contain one line and one space:
- A and B because the 2 requires one more line.
- B and C because the known green line must go left or right.
- C and D because the known blue line must go left or up, (or simply because the 1 requires one line).
- Etc.



So we end up with an alternating sequence of lines and spaces, where A, C, E and G must all be one type (lines or spaces), and B, D and F are all the other (spaces or lines).
But A and G cannot both be lines because of the known line around the 3, so A,C,E,G must be the spaces, and B,D,F must be the lines.