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


I find something strangely satisfying about a well designed slitherlink puzzle, even one which is oddly coloured and plays with the rules a bit.
So I thought it was about time I tried creating one myself. The first version was a bit hard, so I added plenty of extra digits to make it a bit fairer.

Should be solvable by hand with no guessing/backtracking.

Penpa link.

enter image description here

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    $\begingroup$ Do the green squares actually mean anything for puzzle itself, or are they to be used after we solve the slitherlink? $\endgroup$ Commented Aug 2, 2023 at 21:51
  • $\begingroup$ @newQOpenWid It's just a slitherlink puzzle. You could try solving it without the green squares, but you'll struggle to find a unique solution. $\endgroup$
    – fljx
    Commented Aug 2, 2023 at 22:01
  • $\begingroup$ Without the green squares, I can easily find multiple solutions that all work. My question is on if we use the green squares to find a unique solution for the slitherlink (and if so, in what way? I personally am not a big fan of logical deduction puzzles where one has to find the rules themselves). $\endgroup$ Commented Aug 2, 2023 at 22:05
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    $\begingroup$ @newQOpenWid that's the puzzle here, though. To discover the clue leading on to the meaning of the green squares. That's why there is the tag [enigmatic-puzzle] $\endgroup$
    – justhalf
    Commented Aug 3, 2023 at 13:16

1 Answer 1

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The title suggests

The oddly-numbered squares are colored, and the few numbers we have support that theory. Also supporting the theory is the fact that about half of the squares are colored.

This means the colored squares are 1's or 3's, the white squares are 0's or 2's.
The few numbers we have force a few of the first lines, and from there I just followed the slitherlink clockwise around the grid filling in numbers as I went.
Some rules I used:
If the line bordered a white square, it had to be a 2.
If three lines couldn't fit around a colored square, it had to be a 1. This is my first ever attempt at a slitherlink so hope I didn't miss anything! It was fun!
link

The starting point:

The point between the two 3's must have a line going to it from each of the "3" boxes, so these two edges are forced no matter what:
start
Since the box to the left of the "0" is white, it must be even. It's bottom edge must be a line, as the line connecting to the lower "3" can't head in the direction of the "0". This means that this white box must be a "2", and further, that one of the two edges shown in red here must exist:
step2
Looking at the white box above the top "3", it must be even and cannot be a "0". Therefore it is a "2", and the only valid edge remaining is the top edge, which must be a line. From this point on, following that top line to the left and right around the grid is straightforward, as the edges of the puzzle constrain it significantly (versus trying to solve starting from the interior of the puzzle).
step3

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  • $\begingroup$ Notice that the bottom-right 3 is in the corner from a 0 -that means there is a line at the top as well. $\endgroup$ Commented Aug 2, 2023 at 22:23
  • $\begingroup$ Well done, I clearly made this a bit too easy. I'll have to come up with something trickier. $\endgroup$
    – fljx
    Commented Aug 3, 2023 at 8:18
  • $\begingroup$ The top edge of the top 2 is not directly forced, right? It could be the right edge (of course, continuing a bit, we can see it's not possible, but not that direct) $\endgroup$
    – justhalf
    Commented Aug 3, 2023 at 13:18
  • $\begingroup$ @justhalf It can't be a right edge as that immediately violates the white cell to the left of the '0', since we know that is a '2'. Maybe I should have stated that. $\endgroup$
    – Amoz
    Commented Aug 3, 2023 at 14:02
  • $\begingroup$ Ah yea, that's another way to deduce it, but yes, I suppose clarifying it is better. $\endgroup$
    – justhalf
    Commented Aug 3, 2023 at 15:11

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