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A curious Slitherlink Deduction

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*

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).  
>!  
>! [![Cropped slitherlink grid with cells coloured and annotated][1]][1]
>! 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.  
>! [![Cropped slitherlink grid with circled region][2]][2]


  [1]: https://i.sstatic.net/V0ZVQPlt.png
  [2]: https://i.sstatic.net/UDBYL8AE.png

Edit Summary*

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$\begingroup$ Ah thanks! I missed that loop through the 2. $\endgroup$

justhalf
– justhalf

2025-10-13 10:51:06 +00:00
Commented yesterday
$\begingroup$ This certainly works but makes use of a lot more of the board. With trial and error one can already deduce the contradiction in the 2x2 box by only considering the box and the 4 edges going into it. But I don't know whether there is any deduction to apply here that is more general than this specific configuration. $\endgroup$

quarague
– quarague

2025-10-14 06:32:30 +00:00
Commented 23 hours ago
$\begingroup$ @quarague The only additional area used in this case is the immediately adjacent ring of cella. (I may have coloured a few more in that image, but they don't actually assist the solution.) Those adjacent cells have to be included in any deduction you make here, because the lines entering/leaving the 2x2 affect which internal configurations are valid. $\endgroup$

fljx
– fljx

2025-10-14 06:45:23 +00:00
Commented 23 hours ago
$\begingroup$ If you only keep all the lines in the 4x3 block surrounding the 2x2 block and leave the rest of the grid empty, then I think trial-and-error still allows to deduce the edge on the 1 but your deduction wouldn't work anymore. $\endgroup$

quarague
– quarague

2025-10-14 06:48:45 +00:00
Commented 23 hours ago

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