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Here's an odd little sliding puzzle:

enter image description here

As you can see, there is currently one free space, in the centre circle. That circle can spin freely, even when it is holding a piece in its 'holder'. Note, however, that if a piece gets moved around via the spinner, it's orientation also will change.

My question, of which the answer I am not sure of yet:

Is it possible to change the order of the numbers such that every tile is still in the correct orientation?

Edit: @somebody has pointed out a very simple construction!

Thus, there remains the question of WHAT orders are possible. I'm suspecting there is some even/odd permutation argument.

For clarification: The intent is to have all numbers having 4 distinct orientations, none of which can be mistaken for another. I did attempt to draw all numbers asymmetrically, and the 6 and 9 distinctly. As of now, I would like the problem as stated to be more straightforward in terms of its mathematical conditions.

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  • $\begingroup$ I think so? Put 5 in -> right -> make space -> put in, then 6 will be at the top hole, put 6 in, left -> make space -> put in, then move them back in the opposite order so the final order is 123465789 $\endgroup$ – somebody May 26 '17 at 3:43
  • $\begingroup$ That actually ends up with 123564789... interesting! $\endgroup$ – TheGreatEscaper May 26 '17 at 3:50
  • $\begingroup$ Don't have time ATM but I can try to make an interactive version like in a few hours $\endgroup$ – somebody May 26 '17 at 4:02
  • $\begingroup$ But that solution will have the 5 lying on its side. I wonder whether there's a trick so that the nine and six can be on their heads. You can make a configuration with two sixes by moving the 9 to the left via the wheel. $\endgroup$ – M Oehm May 26 '17 at 6:15
  • $\begingroup$ This really depends on rotations, and how the numbers are drawn. For instance, if I rotate 1 180 degrees will you say it's still a 1 or not? As you've drawn it probably no. This needs to be clarified for 8, 9, 6, 5 and 1. $\endgroup$ – Beastly Gerbil May 26 '17 at 6:59
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Mathematically speaking, it should be simple to note that

Even ignoring orientation, all of the possible moves (left-to-middle, middle-to-right, left-to-right or vice-versa) are odd cycles, and therefore even permutations. This means that no odd permutation is constructible, as it can't be formed by a composition of even permutations.

On the other hand,

somebody showed a way to make a 3-cycle:

Put 5 in -> right -> make space -> put in, then 6 will be at the top hole, put 6 in, left -> make space -> put in, then move them back in the opposite order so the final order is 123564789

Since you can rotate the board arbitrarily, you can use this 3-cycle on any 3 consecutive numbers: for example to cycle 6 7 8, do left-to-right twice, cycle, right-to-left twice. As the consecutive 3-cycles generate the set of even permutations, this means that you can generate any desired even permutation by doing the aforementioned sequence some number of times.

Jaap Scherphuis notes in the comments that:

Since you can rotate a single tile in one move and permute the tiles arbitrarily without changing the orientation of any of the tiles, this means any combination of tile orientations is possible for all formable permutations.

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  • $\begingroup$ IIRC the set of even permutations is generated by the set of all 3-cycles, but we have only one 3-cycle so far. Am I missing something? $\endgroup$ – Ankoganit May 26 '17 at 14:14
  • $\begingroup$ @Ankoganit that's what the parenthetical remark was supposed to address -- say you want the cycle 6 7 8, do left-to-right twice, cycle, right-to-left twice. $\endgroup$ – ffao May 26 '17 at 14:25
  • $\begingroup$ What you are using here is that the set of 3-cycles of adjacent tiles, i.e. of the form (k k+1 k+2), generate all even permutations. This is true, because with a bit of conjugation you can construct all 3-cycles of the form (1 2 k), and from those it is fairly easy to construct any 3-cycle (a b c). $\endgroup$ – Jaap Scherphuis May 27 '17 at 5:49
  • $\begingroup$ It should also be noted that you can twist a single tile (just do one move). Combining that fact with the fact that any even permutation is possible without changing the orientations, means that all tile orientations are possible regardless of the permutation. The total number of states of the tiles (assuming the spinner is empty) is therefore $4^9*9!/2 = 47,563,407,360$. $\endgroup$ – Jaap Scherphuis May 27 '17 at 5:57
  • $\begingroup$ @JaapScherphuis of course, I was thinking about orientations but missed this very simple proof, thanks :) $\endgroup$ – ffao May 28 '17 at 9:22

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