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enter image description here

Forming a 2x2x1 puzzle block, four 1x1x1 Rubik's cubes were modified to attach with one another in many ways by means of magnets and steel discs.

Assemble them first so that all sides are showing the same individual cube face colors. Shuffle by twisting (90 or 180 degrees-CW/CCW) any two side by side cubes on the axis where both their centers connect. Just enough to make it challenging. Solve or play it. Same rule of rotation as shuffling shall be applied on 4-axes as shown. The solved state may not be like the initial same side color configuration. You can reshuffle to play again with fewer moves.

If by then only one of the cube is disoriented, how many turns more shall it take to complete the puzzle?

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    $\begingroup$ I don't quite understand how the moves work; is it the same way as a 'regular' 2x2x1 cube? What does the diagram in the lower-right of the picture mean? What is the solved state if it's not "like" the initial state? $\endgroup$
    – 2012rcampion
    Oct 30, 2016 at 23:19
  • $\begingroup$ Not like 2x2x1, this is made of 4 dettachable 1x1x1 but would always stick together to form the 2x2x1. To turn, dettach 2 side by side cube together ,rotate it , stick it back on. $\endgroup$
    – TSLF
    Oct 31, 2016 at 4:30
  • $\begingroup$ The diagram is the 2 magnets and 4 steel discs on 6 faces of all cubes beneath the stickers. Note that a cube with this mechanism would stick to another cube's face in 24 ways (aligned sides) $\endgroup$
    – TSLF
    Oct 31, 2016 at 4:35
  • $\begingroup$ Can we rotate it in any way we want, or are we limited to the generic 180 degree rotation? Like, can we roll the 2x1x1 90 degrees? $\endgroup$
    – greenturtle3141
    Oct 31, 2016 at 4:38
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    $\begingroup$ Can I clarify the question? You want to get to a state, via a sequence of legal moves, where only one cube is disoriented with respect to the others and then solve that in the minimum number of moves, is that correct? Are we looking for a minimum solve? $\endgroup$
    – hexomino
    Oct 31, 2016 at 9:54

1 Answer 1

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I think from the state of one disoriented puzzle block, the minimum number of moves required to solve is

$4$

To see this

We will start with a solved puzzle block and perform a sequence of $4$ moves so that the result will be that just one of the cubes is disoriented. The solution is achieved by simply reversing this procedure.

First let us consider the solved state as shown in the diagram (sorry about the mess) enter image description here

For our purposes, we will say that, for each individual cube, the face opposite red is orange, the face opposite blue is green and the face opposite yellow is purple. Now, if I understand correctly, a single move involves considering one of the axes $A$, $B$, $C$ or $D$ and rotating the two cubes which intersect that axis clockwise about it by an angle of either $\frac{\pi}{2}$, $\pi$ or $\frac{3\pi}{2}$.

Here, we will set the convention that to perform a move you stand at the position of one of the letters as in the diagram and perform the rotation for the move. This will give us the sense of what "clockwise" means for each axis.

From the solved state, perform the following four moves:

Rotate $A$ by $\frac{\pi}{2}$.
Rotate $B$ by $\frac{\pi}{2}$.
Rotate $A$ by $\frac{3\pi}{2}$.
Rotate $B$ by $\frac{3\pi}{2}$.

The result is the following: enter image description here
Essentially, ever cube is preserved except the bottom right hand corner. To solve this we can perform the reverse sequence of moves:

Rotate $B$ by $\frac{\pi}{2}$.
Rotate $A$ by $\frac{\pi}{2}$.
Rotate $B$ by $\frac{3\pi}{2}$.
Rotate $A$ by $\frac{3\pi}{2}$.

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  • $\begingroup$ How many disoriented 1 cube positions are possible? Can B-A-B-A solve them all? $\endgroup$
    – TSLF
    Nov 2, 2016 at 16:47
  • $\begingroup$ That's a good question. I'll think about it. One more obvious one you can get is performing the above sequence of moves twice, which gives you purple on top, blue on front and orange on side but we should be able to get more by switching things round a bit. $\endgroup$
    – hexomino
    Nov 2, 2016 at 17:04

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