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Solve 5 Slitherlinks, every with a novel answer.
Please say a phrase concerning (or present a picture depicting) every answer methodology.
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I checked out different folks’s solutions for the primary 4 grids, and received the options from these. However, this is what I feel the fifth grid is (minor spoilers for the opposite 4 grids):
Place the 4 tetronimoes from the options to the primary 4 grids collectively right into a 2×2 sq., with out rotating them. There’s just one approach to do it:
+-+-----+
|?|? 2 ?|
| +-+-+ |
|1 3|3|?|
| +-+ +-+
|?|?|2 ?|
+-+ +-+ |
|3 ? ?|?|
+-----+-+
It has a novel answer:
..+-+-+-+
.?|?.2.?|
..+.+-+.+
.1|3|3|?|
..+-+.+.+
.?.?.2|?|
+-+-+-+.+
|3.?.?.?|
+-+-+-+-+
This is copied instantly from my notes; I used to be utilizing+
for an intersection the road undoubtedly went via, and.
for an intersection or edge it undoubtedly did not undergo. The predominant remark is that the one approach to get a line via a pair of 3s is to zigzag it previous; in case you zigzag it the opposite approach, the two forces it to get caught within the top-left nook, so the one doable path for the road goes spherical like this.
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I’ve discovered 4 slitherlink options, however I’m not sure learn how to assemble the fifth…
I initially thought the gadgets within the tetrominoes had been potential values for that sq. in a 2×2 Slitherlink, however there isn’t any approach to make that work with out rearranging the squares. (ex. 3/3 on the proper aspect is a contradiction, and three/1 would pressure a 1 on the backside left, which solely has choices for 3 or 2.)
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