As we know, 3's seems to be very rare in slitherlinks. This is the slitherlink with only 3's.
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3$\begingroup$ 3's are quite easy to deal with once you know the patterns. This was fun to try to solve very fast. $\endgroup$– ascheplerNov 25, 2020 at 4:09
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$\begingroup$ I haven't had a chance to study this puzzle. In the meantime, I pose this question: is this the Slitherlink with only 3's or a Slitherlink with only 3's? $\endgroup$– Brian DrakeNov 25, 2020 at 14:49
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3$\begingroup$ @BrianDrake Technically it should be a, since we can have very many different slitherlinks with only 3's. $\endgroup$– BubblerNov 26, 2020 at 0:09
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2 Answers
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Here are some tips for working on 3's in Slitherlink:
- Whenever two 3's are adjacent, you can draw some lines sandwiching them.
- Also, the line between two 3's cannot be connected outside (it must connect to top/bottom layer).
- Whenever two 3's are diagonally touch, you can draw some lines encircling them.
- And whenever 3 is cornered, draw lines to that corner.
And with some basic deductions (all lines must be connected in a single loop), here is the solution:
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$\begingroup$ Why couldn't the bottom-right lines of your example be lines? They can't both be lines, but what stops the line from going along the bottom of the grid, up and around the 3, back down to the bottom, then up along the right side of the grid? $\endgroup$ Nov 24, 2020 at 17:18
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2$\begingroup$ @user253751 You're misinterpreting that. What it means is in the case where we already know the bottom-right lines are not filled, then we can deduce that the other two are. $\endgroup$ Nov 24, 2020 at 17:23
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Firstly, we can make the following deductions:
If there are adjacent 3s horizontally or vertically, then their horizontal or vertical components must be part of the loop respectively. If there are diagonal 3s, then the segments which are furthest away from each 3 must be part of the loop. Based on this, we can make some quick progress since there are a lot of 3s adjacent or diagonal to one another.
Next, focus on the 3 in R15C9. If the top part of that three were to be part of the loop, then the right side of that 3 will become trapped. Hence, the bottom part of that cell must be part of that loop.
A similar deduction can be made for the 3 in R14C1. If the bottom part of that 3 was to be part of the loop, the left side of that 3 will be trapped. Therefore, the top part of that 3 must be linked to the loop.
The last deduction we can make is for the 3 in the corner at R15C15. It has only 1 way to be linked to the rest of the loop. Using the deductions so far, we resolve the bottom part of the grid:
Next, focusing on the 3 in R1C1, the loop segment there has only 1 way to link up with the rest of the loop. This allows us to finish the top row.
Then, focusing on the right side: if the cell in R9C13 had its top part as a segment of the loop, the left side will become trapped. Therefore, the bottom part of that 3 must be a part of the loop instead. The chain deductions resolve the right side of the grid:
If the cell in R8C11 were to have its right side as part of the loop, it will result in an unconnected loop. Therefore, its left side must be connected instead. This leads us to
Now, the cell in R3C9's right side cannot extend downwards and neither can the cell in R5C11's top side. Therefore, these two loop segments must connect to each other
After that, if the bottom side of the 3 in cell R5C9 was connected, then it will result in a broken loop. Therefore, its top side must be connected. After that, there is only one way to connect to the 3 in cell R8C9.
Finally, the left side of the grid can be resolved trivially. This lets us resolve the grid
