$\mbox{R}_2\mbox{U}_2$ Back To The Future

The $2\times2\times2$ cube has 8 moving parts, the 8 "cubies" at each corner. The cubies can be divided into two layers (top and bottom) and are easily mapped onto a grid. From here on each cubie is represented by an integer $x \in \{1,\ldots,8\}$ .

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In this representation, we can clearly see the effect of each of the moves $\mbox{R}_2$ , and $\mbox{U}_2$ :

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As indicated by the grey boxes, cubies 7 and 8 are never permuted under $\mbox{R}_2$ or $\mbox{U}_2$ , they stay fixed always.

The first permutation, $\mbox{R}_2$ , permutes cubies 1, 2, 5, and 6. Specifically, 1 swaps with 6, and 2 swaps with 5. We can say that $\mbox{R}_2=\sigma_{\mbox{R}_2}$ given by $\left(1\mbox{ } 6\right)\left(2\mbox{ }5\right)$ .

Similarly, the second permutation, $\mbox{U}_2$ , permutes cubies 1, 2, 3 and 4. Specifically, 1 swaps with 3, and 2 swaps with 4. We can say that $\mbox{U}_2=\sigma_{\mbox{U}_2}$ given by $\left(1\mbox{ } 3\right)\left(2\mbox{ }4\right)$ .

What this notation means is that any numbers enclosed by a set of parenthesees map into one another (left to right) under the permutation. For example, $\left(1\mbox{ } 2\mbox{} 3\right)$ means that 1 maps into 2, 2 maps into 3, and 3 maps into 1, under the permutation.

In general, we can represent the collection of all moves on the $2\times2\times2$ as a set $\mathcal{R}$ endowed with a composition function $*$ such that $\forall \mathcal{R}_1,\mathcal{R}_2 \in \mathcal{R}$ , we have $\mathcal{R}_1*\mathcal{R}_2 \rightarrow \mathcal{R}_3 \in \mathcal{R}$ where $\mathcal{R}_3$ is also in $\mathcal{R}$ .

In other words, applying two valid cube moves in sequence results in another valid cube move. In this way, $\mathcal{R}$ along with the function $*$ form an algebraic group.

The permutation we're interested in, $\mbox{R}_2\mbox{U}_2$ , is given by the composition $\sigma_{\mbox{R}_2}*\sigma_{\mbox{U}_2}$ . We can form the composition of the two permutations by following the trajectory of each element as it is shuffled by each. If we find that there is a subset of $n$ elements that form a closed loop under the two permutations, i.e. $a\rightarrow b\rightarrow c \ldots z\rightarrow a$ , we say they they form an $n$ -cycle. In general, a permutation is made up of a union of disjoint cycles.

Let's find the permutation that represents $\mbox{R}_2\mbox{U}_2$ :

$\begin{aligned} \sigma_{\mbox{R}_2\mbox{U}_2}&=\sigma_{\mbox{R}_2}*\sigma_{\mbox{U}_2} \\ &= \left(1\mbox{ } 6\right)\left(2\mbox{ }5\right) * \left(1\mbox{ } 3\right)\left(2\mbox{ }4\right) \\ &= \left(1\mbox{ } 6\right)\left(1\mbox{ } 3\right)\left(2\mbox{ }5\right)\left(2\mbox{ }4\right) \\ &= \left(1\mbox{ } 6 \mbox{ }3\right)\left(2\mbox{ }5\mbox{ }4\right) \end{aligned}$

So $\sigma_{\mbox{R}_2\mbox{U}_2}$ is composed of two disjoint 3-cycles, one that cycles $\{1,3,6\}$ and one that cycles $\{2,4,5\}$ .

If a permutation is composed of multiple disjoint cycles, the permutation will repeat itself after it is applied $\gamma$ times, where $\gamma$ is the least common multiple of the component cycle lengths. In this case the component cycles both have length 3 so the permutation $\sigma_{\mbox{R}_2\mbox{U}_2}$ will return to the original state after 3 times. I.e. $\sigma_{\mbox{R}_2\mbox{U}_2}^3 = e$ where $e$ is the identity element of the group, i.e. $x*e\rightarrow x \quad \forall x\in \mathcal{R}$ .

To learn more, consider reading the group theory article in the techniques section.

A way to solve this, although time consuming if applied to larger cubes, would be to number the 8 cubes 000, 001,... 111 as if they were 3D coordinates (xyz) in that order. From observation, we can see that 000, and 001 do not move. Let the remaining 6 cubes be numbered 1 to 6: 010 = 1; 011 =2;... 111 = 6.

The operation R2U2 maps the positions as follows:

1 --> 3

2 --> 4

3 --> 6

4 --> 5

5 --> 2

6 --> 1

where 1 --> 3 denotes that the cube in position 1 (010) is now in position 3 (100). After 3 cycles, the cube is back to the original position.

1 --> 3 --> 6 --> 1

2 --> 4 --> 5 --> 2

3 --> 6 --> 1 --> 3

4 --> 5 --> 2 --> 4

5 --> 2 --> 4 --> 5

6 --> 1 --> 3 --> 6

*note how two sets of numbers cycle ( 1 --> 3 -->6) and (2 --> 4 --> 5) please comment if there is an explanation of this without using group theory

Define the "parity" of the cube $(x,y,z)$ as the parity of $(x+y+z)$ , and note that $R_2$ and $U_2$ both preserve parity.

Considering the effect of $R_2 U_2$ on the four even cubes, we see that it leaves one alone and changes all of the other three. But that necessarily means that it will take exactly three iterations to bring all even cubes to their original states. By identical reasoning, it will take exactly three iterations to bring all four odd cubes to their original states. Therefore, the answer is three.

I just did it on my $\huge ~~Cube~~$