As far as I know, the only way to figure this out is by letting a computer run through all the possibilities. It is a small puzzle, so this does not take long.
First I will assume that you want the final solved position to have the blank in the bottom right corner, with the tiles in numerical order:
(See further below for the results with the ...
As answered by @BeastlyGerbil, you have the world of twisty puzzles. Here in the Twisty Puzzle Museum you can find over 5,000 of these kind of puzzles, and here is my personal collection of currently 279 puzzles (pictures are a bit outdated though, since I now have a few more shelves; list is up-to-date however).
That being said, you gave the following ...
Kevin Cruijssen gives some good examples. You might also want to consider multiple layers as a part of the puzzle, which allows you to have additional constraints, either visible or hidden.
For example, "One Fish, Another Fish", where the frame and piece shapes constrain movement
I highly recommend looking at Rob's Puzzle Page ( http://robspuzzlepage.com )...
Here's my attempt (maybe I'll get graphical when I have the time, text only for now.)
First, label the non-dynamite barrels like this:
and develop a notation: <Barrel name>-<direction> means move the named barrel to the mentioned direction, by whichever means it can move there. (It will always be unique.)
Then, make the following moves:
Sacks 2 & 3
Firstly, since I did this using a grid and its hard to rotate the diagrams, they've been rotated 45 degrees.
It's quite trivial to get everything else out as all moves are reversible so in the worst case scenario move everything back to the beginning and then put purple where green was.
Not required to do it in the minimum number of moves, eh? Well, "Inefficient" is my middle name! Actually, it's 'Pink', but whatever.
Let's build a horribly inefficient solution by first making some useful-but-inefficient building blocks, and then strategically-but-inefficiently combining them together!
Based on the moves I ended up creating, my strategy ...
The hardest positions on the 15-puzzle require $80$ moves to solve (where a move consists of sliding a single tile). Here is an example of a position requiring $80$ moves:
This position (and ...
The standard way of determining the solvability of sliding-block puzzle is analysing the parity of the corresponding permutation. A permutation is simply a reordering of some things — the tiles, in this case. We can assign a notion of parity – i.e., evenness or oddness to every permutation.
To understand how this works, consider the following ...
My solution has:
I will use P to denote pawn for nice formatting (sorry chess fans) and instead of writing multi-move chains of sliding the same type of piece in a line I will just put the start of the chain, the end of the chain and the number of moves between in brackets.
The key part is the start:
This is the brute force part:
This gives ...
I'm sure you know of the Rubik's Cube. That has got multiple movable pieces.
If you want something with a lot more movable pieces you could try some of the family members of the Rubik's Cube.
For instance the Gigaminx:
Or even the Teraminx:
There are weirdly shaped ones like the Ghost Cube:
Or Fisher Cube:
And even jointed ones like the Siamese Cube:
OK, this one's going to be way more difficult than the ones before. The first observation is that an open barrel standing somewhere can only stand somewhere else if it has rolled in between, and rolling requires a two-space wide path. And the doubly open barrel of course cannot even stand at all. So, let's get rolling! (Hlyaaargh.)
Since the door isn't wide ...
For larger versions of the n-puzzle, finding a solution is easy, but the problem of finding the shortest solution is NP-hard. For the 15-puzzle, lengths of optimal solutions range from 0 to 80 single-tile moves or 43 multi-tile moves.
And for the 24-puzzle:
In 2011, a lower bound of 152 single-tile moves had been established; current established upper ...
For a 15 puzzle to be solvable it has to meet the following:
If the grid width is odd, then the number of inversions in a solvable situation is even.
If the grid width is even, and the blank is on an even row counting from the bottom (second-last, fourth-last etc), then the number of inversions in a solvable situation is odd.
If the grid width is even, and ...
As mentioned before. If you've done these kind of puzzles before a common strategy is to solve row by row but the final two rows together just because it won't always work out.
Consider this bottom right segment of any puzzle larger than 2x4 in correct order:
you can now do this to make the top row still correct but the bottom not.
The same argument for the $15$ puzzle applies. You need the starting position to be an even permutation, so $1/2$ of the starting positions are solvable. You follow the same proof that shows you can swap two pairs of numbers, but not one pair.