3
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Here's an instance of a $2$x$3$ sliding puzzle:

Imgur

The $2$x$3$ sliding puzzle is just a smaller version of the well-known 15 puzzle. Many guides exist on solving the 15 puzzle, and there are strategies that work on any $n$ x $m$ puzzle, where $n \ge 3$ and $m \ge 3$, such as the one in this video. But these solutions I've found don't work in the cramped $2$x$3$ puzzle, or more generally, in a $2$x$m$ puzzle.

What strategies can be applied to solve these types of puzzles?

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2
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If the puzzle is solvable, then what you have to do is just putting 1 2 on the top (then the rests will automatically be solved).

Somehow, you managed to put 2 1 instead. In this case, one of possible way is to divide 1 and 2 like this:

1 X
2 _
X X

and

1 X
X _
2 X

Then rotate 4 tiles on above then below:

X 1
X _
2 X

and

X 1
X _
X 2

Then voila, just rotate them all so 1 2 will be on top!

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  • 1
    $\begingroup$ This can be turned into a strategy for any 2-by-m puzzle. For each row, from top to bottom: get the two tiles that belong in that row, in that row (not necessarily in the right order). That should be straightforward. Then, if they need to be swapped, apply the given transformation. $\endgroup$ – Seeker14491 Mar 11 '18 at 22:15
4
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A solution:

21
x4
53

14
23
x5

14
35
2x

x4
15
32

45
12
x3

45
23
1x

x4
25
13

24
15
3x

12
34
5x

The first thing to note is that 21 and 43 both need to be swapped, so the parity is okay. To split a pair in this puzzle, you need to get them top and bottom, otherwise they are not independent.

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