# Tag Info

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The question is a bit unclear - it sounds like you're asking "Given a sequence of moves A, is there always some number of repetitions of A that will get you back to the start state?" This is relatively easy to prove - it's just basic group theory. Imagine you have a bunch of Rubik's cubes -- one in every possible state. Given an algorithm A, you ...

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They all

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It seems like you are trying to find One such example that satisfies this is In particular, As mentioned by armb in the comments, there is a good answer here discussing the maximum orders for an $n \times n \times n$ Rubiks cube.

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The password is First, solve the Rubik's Cube. Now to find the password.

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I believe the answer is First, note that... Then...

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This is the superflip, the position of a 3x3x3 cube where all edge pieces have been flipped in place, but applied to a 5x5x5 cube using only its outer layers. My preferred method for doing this is the following: (Shown for a 3x3x3 cube, on a 5x5x5 treat the inner layers as if they form a single middle layer)

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The cube is definitely defeatable if The way to beat the cube is to If this method is considered too trivial and against the spirit of the rules, you can still do it for by making use the non-trivial identity

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I think Like so: Since the cubes are all similar, it's sufficient to check that 2 of the colours end up correctly oriented, the order of the other colours is then forced. To avoid collisions, the top left cube can be moved first (all but the last square), then the top right and bottom right cubes can go to their spots, followed by the top left cube's final ...

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This seems to be the 8355 method by Reheart Sheu. It seems the original blog was Taiwanese but no longer exists, but the speedsolving wiki has a description. Step 4 consists of solving 3 of the last layer edges intuitively (U face, yellow edges), and then the last two edges (one in the U layer, and the keyhole edge in the middle layer at FR) may need to be ...

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This is what Glenn said on JRCuber: "I have only one Stickerless 3×3, 6 Stickered. To me there is a completely different feel you can go way faster if set up properly. As well as you don’t have to worry about nails scratching and stickers grinding. All smooth plastic. Plastic for speed solving. Stickered for big cubes. Hope that helps."

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It is possible to swap two corners, but doing so will also swap two edge pieces. It is possible that there are two edge pieces on the cube that are identical, and swapping those two will not be visible, so that you can in effect swap two corners without seeming to have any effect on the edge pieces. Notice that the centre square showing a 5 is twisted by a ...

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Then

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The Gear Octahedron puzzle actually has $1,327,104$ states. On my website I have a page about the Gear Mastermorphix, which is an equivalent puzzle (same mechanism, but with a tetrahedral outer shape). On that page there is an explanation as to how that number comes about: $4!$ permutations of the triangular face centres (one of them is held steady and ...

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Several people commented that is mechanically identical to a regular 3x3x3 Rubiks cube. Assuming this to be correct, this imagined re-stickering of the cube could perhaps let you visualise how it is identical: The pieces that would normally be edge pieces sharing an edge with the white face are unusually large (reaching to the actual corners of this cube), ...

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You can cycle three centre pieces of the top face as follows, using small letters to denote turns of inner layers: r' d2 r (this removes from the top face the front-right centre piece) U (this brings a different top face centre piece to the front-right) r' d2 r (this swaps the front-right centre piece with the previous one) U (this brings a different top ...

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It seems that this problem needs an additional assumption for the initial configuration of the cube at the start of the game. If the initial cube configuration is not predefined (i.e. random initialization), then there doesn't seem to be a way to know if you are "The first prisoner to enter the room" (Callidus' solution). Similarly, for any other ...

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I want to build on ghosts_in_the_code and Jens Renders' ideas and provide a practical solution easy to perform and to memorize.

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Sure. You should start with a list of the small cubies (red-blue-white corner, green-yellow side, orange center, etc), and make a random configuration with the one required side in place. Then, you must guard against the three possible parity errors that can cause an invalid scramble: The total amount of twisting that the corners need must be sum up to ...

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This is in principle exactly the same as just solving the cube, but in practice it's much harder on the brain. Here's a way to do it that's kinda cheaty but demonstrates the equivalence. You're starting with a solved cube. For each facelet[1] of the cube, figure out where it needs to go. Look at what colour is in that place right now, on the solved cube. ...

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Set up the pattern you want. Solve it, writing down the moves you used. You now have a solved cube. If you undo the move sequence you wrote down, you will get the original pattern. In other words, if you invert the written move sequence (reverse the order of the sequence, and the turning direction of each move) you get a move sequence that generates the ...

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If you have a standard Rubik's cube, and draw arrows on each label (all arrows on each face point in the same direction), is it possible to solve the cube such that some of the arrows face different directions from the original orientation? Yes, this is possible. The centres can be rotated. To rotate the U-face centre by 180 degrees, you can do the move ...

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This page tells which algorithms one can look into. Bear in mind that you will have to rotate at least two centers at a time.

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There is no way to do a 4-cycle of edges without affecting anything else, because only permutations with an even parity can be performed. Note however that on your cube the centres are also incorrect. If you shift the middle layer to the right one quarter turn to put its centres correct, then its edge pieces become the incorrect ones. You then have 8 ...

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The difference is that the Numerical IPG v5 actually comes with 4 extra sets of tension nuts. [Numerical IPG] Equipped with 4 sets of GTN which are explained (much) later down the page in the section 'The new GES+'.

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Stickerless cubes used to be banned because for a while mostly because there was the idea that you could cheat if the cube was partly turned. If one of the layers is at a 45 degree angle, then you can see both the color of the 'sticker' facing you and the 'sticker' on the other side that you wouldn't normally be able to see. Doing this 45 degree turn on a ...

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You are wrong in assuming you need to do a 1/4 turn of the center. Turning the 4 pieces of the center by 90° is impossible. It amounts to an odd permutation of the centers, which is impossible to do because all moves on an 4x4x4 do even permutations on the centers. But you will notice that the centers are not all different. On the picture you see that the ...

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OP confirmed it was a regular 3x3x3 case, where three of the edges have to rotate around without orienting them. In that case, the following algorithm could be used with the solved fourth edge at the top-front:

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It is rather difficult to extract this information directly from the current locations of the corner pieces. By far the easiest way is to actually try to solve those corner pieces using only half turns (while ignoring the edge pieces), and see how far you get. For now I'll assume that the corner pieces are already located in their correct tetrad orbits {UFR, ...

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Speedsolving.com user rjohnson_8ball has figured out a way to reorient 2 of the centre pieces on a single side (while reordering them a bit, which shouldn't be an issue here), so running his algorithm twice (properly orienting the cube each time, of course) should do the trick: https://www.speedsolving.com/threads/4x4-picture-cube.5181/post-68020 I'm copying ...

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