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The puzzle here is to analyze the following Slitherlink configuration and show that it is impossible.

.------------------
| .   .   .   .   .
|           2
| .   .   .   .   .
|       2
| .   .   .   .   .
|   2
| .   .   .   .   .
| x
| .   .   .   .   .

The x-ed edge means that it is assumed that no line can be placed in the edge. Please avoid as much as possible a lengthy case by case analysis, or a vague explanation.

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1 Answer 1

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There are only two scenarios that can arise for the bottom 2 (that don’t immediately lead to an unsolvable grid) let’s take a look for each one and see how they both lead to dead ends meaning the puzzle is unsolvable.

A)

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With the bottom 2 lines being configured this way this result is forced and the joining of any two of the 3 paths as you can see leads to the impossibility of the third path being connected as part of a loop and thus being stranded.

B)

enter image description here

enter image description here

The same goes for the second arrangement as you can see this result is forced with the path of the bottom 2 being configured this way but this again leads to the same problem as above.

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  • 4
    $\begingroup$ There are actually four ways to place the lines around the 2's, you only have two of them here. The key point here is that lines must exit a 2 square from diagonally opposite corners (either a single pair, or both pairs). The forbidden line blocks one pair for all three 2's, so they must all have a line leaving from the top-left corner. That means you end up with an odd number of ends in that corner no matter where the actual edges go around the 2's. $\endgroup$
    – fljx
    Commented Feb 12 at 20:20
  • $\begingroup$ I am fully aware there are four ways I omitted 2 of them them because they were obviously leading to an unsolvable grid $\endgroup$
    – PDT
    Commented Feb 12 at 20:53
  • $\begingroup$ I probably need to explain it more clearly $\endgroup$
    – PDT
    Commented Feb 12 at 20:55
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    $\begingroup$ I'm not talking about just the bottom 2. Four ways in total of placing edges around all the 2's, given the forced no-edge at the bottom. There are two other ways to place the lines around the upper 2's in your case B. There is nothing that makes those other two arrangements more or less obvious than the one you have chosen. $\endgroup$
    – fljx
    Commented Feb 12 at 22:15
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    $\begingroup$ @fljx but for any of those 4, the reasoning to reject them is the same, so I don't think it's that important to point out the exact number. Just like you said, the key point is that it'll exit at diagonally opposite corners, so this configuration will always end up with odd number of end points in the corner, impossibility. $\endgroup$
    – justhalf
    Commented Feb 14 at 3:47

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