16

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.


8

For a scramble to be any good, at the very least you would want every pair of adjacent pieces to become separated at least once during the scramble. You don't want any blocks to remain unscrambled, because that would make it slightly too easy. Here is a very rough calculation of how many moves this takes. There are $48$ pairs of adjacent pieces ($6$ cutting ...


7

The algorithm is $r \ U^2$ - X - $r \ U^2 \ r\ U^2 \ r' \ U^2\ l \ U^2 \ r' \ U^2\ r \ U^2\ r' \ U^2\ r'$ Where $r$ means you have to turn two layers from the right upwards and $r'$ means downwards, $l$ means you have to turn two layers from the left downwards. and x means you have to turn the whole cube one face upwards. Initial $rU^2$ $\times$ ...


6

There is nothing special about the solved state. Solving the cube means achieving a particular permutation of the cubelets' faces; that's true whether what you're trying to end up with is the state we call "solved" or some other state; God's number is just the maximum number of operations you need to do to achieve any achievable permutation of cubelet faces.


6

It's a 4x4 Parity. The algorithm to solve is, $(2R)^2U^2(2R)^2u^2(2R)^2(2U)^2$ Where $2R$ is the second layer from right, $U$ is the top layer, $u$ is top two layers combined, $2U$ means second layer from top and the superscript implies you have to turn twice. First hold the cube in this position. STEP 1: $(2R)^2$ STEP 2: $U^2$ STEP 3: $(2R)^2$ ...


6

Yes. This is a 3-cycle of the corners, and all three of those corners also need a twist in the same direction. Both of these things can be done individually. I put that position into Cube Explorer, and it gave the move sequence R' F U2 F' R F R' U2 R F'.


6

I think: Also This leaves: Beyond this:


5

Assuming you took the cube apart and are trying to put it back together, the question is how the top layer can all be black on top, but the pieces around the edges don't match. The answer is that the only cube we see all 3 sides of is the piece in front of us. The blue and orange pieces have a colored face that is oriented towards the inside of the cube, ...


5

If the three edge pieces you want to rotate are on the top layer at front, left and right, then the move $F^2UM^{-1}U^2MUF^2$ will permute them cyclically where $F$ means rotating the front face $90^{\circ}$ clockwise (so $F^2$ means rotating it $180^{\circ}$), $U$ means rotating the top (upper) face $90^{\circ}$ clockwise, and $M$ means rotating the ...


4

Here is one method for attaining such a pattern. I found this pattern as follows: EDIT: A shorter method is:


4

If a single corner of a rubik's cube is twisted, it is in an unsolvable state. The only way to fix it is to take it apart and put it back together again ;)


4

Unless the piece at the opposite corner has a similar misorientation, the cube has been tampered with. There can never be a single piece wrong. A set of moves to twist one piece, always has a corresponding side-effect.


4

This takes 7 moves, and cannot be done in less (I checked by computer). Here are all the optimal ways to swap the two UF corners using only moves of the U, R, and F faces: U F U' F' R' F' R U' F' U F R U R' R U' R' U' F' U F R U' R' F' U' F U R' F R U F U' F' R' F R F U F' U' F U F' U' R' F' R F' U' F U R U R'


4

The move notation with small letters is ambiguous, as different sources use this to mean different things. Ruwix's Advanced Notation page explains that they use the small letters to denote a turn of the two outer layers. This notation is often used on the 3x3x3 cube as well. So yes, the site uses u to to mean what you know as Uw. For larger cubes they use $...


4

First, we tabulate all the adjacencies: Now we need to look in a little more detail. We'll start with the corner pieces. I'm going to rotate the center pieces around to make it so that there are fewer moves. We also need to remember the parity rules for a rubik's cube: So now we need to work on the edges! The indexing we'll use for the edges is: Hence ...


3

From your pictures, I found the orientation as below. Keep the red face on top as below, Then follow the below algorithm: L' U' L F - L' U' L U - L F' L$^2$ - U L U L' - turn the left layer up L - turn the left layer down U - turn the top layer clockwise U' - turn the top layer anti-clockwise F - turn the front layer clockwise F' - turn ...


3

The piece at top front is from another puzzle. There are two pieces shown with black/blue on the same faces - the other is on the top left corner. There can't be two black/blue pieces with another colour on their left, in the same puzzle. And to think you didn't even bother replacing the missing piece at the bottom!


3

I don't think this is an invalid state. First you need to complete the two layers starting from green. In the second picture, there are two blocks in the second layer which are not correctly solved. First solve them, and confirm that first and second layers are correct. After this, try the yellow cross step with the blue layer.


3

This cube can often be fixed by just forcing that corner without complete disassembly. Just twist that corner. Sometimes easier if you move that corner half way between two positions.


3

It is not possible to swap a single pair of edges by doing legal cube moves. The simplest way to fix it is, well, to fix it: take it apart enough that you can bodily swap the two edge pieces back again.


3

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 ...


3

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 ...


2

Why so complicated? You can use even a simpler combo. First, move the other flipped edge to the opposite side. If BLUE is Front and ORANGE is Left do: Moving the orange edge to the opposite side: L', B' Combo to remember: 4x(M'U), U, 3x(M'U), M' Moving back the orange edge to where it was: B, L Basically you can flip sides on any two edges anywhere using ...


2

Yes but I was able to shuffle it further without the checker board pattern so it even looked random! This took me about four days because I am bad at puzzles and I haven't touched it since!


2

It can be solved. This seems to be the antisune case which can be solved with the algorithm:


2

I worked through several approaches to solve this and really enjoy the challenge. Given the number of permutations that would satisfy the given conditions, proving that a solution is optimal may require brute force. I decided to do it the hard way, trying combinations of patterns both in Cube Explorer and my own cube. At first I started with leaving as ...


2

Looks like you have a single flipped edge piece. This renders your cube unsolvable: If you ignore everything except the corners, centers, and edge middle pieces, you get a 3x3x3 cube, and on that cube, a single piece flip is definitely unsolvable. Here's how to get from a solved cube to a situation that is one edge piece flip away from your cube's pattern. ...


2

Instead of taking it apart, you could also, optionally do one of six things: Take off and rearrange the stickers, Take it back to the store, Tell your friend/family member/distant relative/creepy (probably a stalker) stranger to stop buying you prank gifts, Throw it away, Buy a new one, or Smash it with a sledge hammer out of frustration and anger, ...


2

It is the Nb-perm that you found but the pdf ignores the edges in that step because it's using 4 look instead. Swapping the edges comes afterwards when all 4 corners are in position. (Except for a final U turn which is skipped until the 4th step)


2

The peeled cube looks like this in its solved state:


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