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Here's a solution (at least the upper bound of 125 moves). I've outlined it, to emphasize the key logic, and created an animation to show the whole solution):

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Using Dr Xorile's answer as a base here is an improved solution:

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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: 123 456 78. (See further below for the results with the ...

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The solution to the crossword is: And this can be arranged as: My strategy: But is this solvable as a 15-puzzle? An explicit solution:

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I think this works:

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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over 5,000 of these kind of puzzles, and here is my personal collection of currently 279 puzzles (pictures are a bit outdated though, since I now have a few more shelves; list is up-to-date however).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 ...

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I hacked together a little solver in Python. The puzzle has a rather large state space, with about 6 million positions. My solver doesn't find any solution shorter than 295 moves; here they are, with apologies for the not-very-elegant notation. Read each row from left to right. Blocks are assigned letters independently in different positions, and the ...

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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 )...

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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: $$\begin{array}{|c|c|c|c|} \hline 15&14&8&12\\\hline 10&11&9&13\\\hline 2&6&5&1\\\hline 3&7&4&\\\hline \end{array}$$ This position (and ...

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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: 1: D-up 2: ...

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Took me some time, but I got 13 steps:

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Sacks 2 & 3 Firstly, since I did this using a grid and its hard to rotate the diagrams, they've been rotated 45 degrees. Sack 2 Sack 3 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.

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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 ...

9

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 ...

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The puzzle is: The steps to show that: Which gives this configuration:

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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 ...

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Bottom Puzzle: Middle Puzzle: Top Puzzle: (AKA: Pain and a half)

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Initial grid as follows:

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My answer: Most questionable word: Note that

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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 gives: This is the brute force part: This gives ...

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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 ...

8

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: So ...

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As @IAmInPLS made clear the letter is This letter reminded me of one logo Thus the animal is

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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 ...

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It is known that, for any plain rectangular sliding puzzle (larger than 2x2), any parity-conforming configuration is reachable from the solved state, and all of those with wrong parity are unreachable. "Parity-conforming" means that the parity of the entire board, plus the Manhattan distance of the hole from its solved position, must be even. ...

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Solution: It can be shown using some basic group theory, similar to the proof found here.

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I had the same problem. Eventually I realized that ... Hope I've helped!

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Since the 'donimoes' are limited to moving along their long axis, I believe this can be done in 12 moves as follows: Visually:

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I have a solution that takes

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