The Rubik's cube was invented in 1974 by Erno Rubik, a Hungarian who originally called this toy, the Magic Cube. His puzzle hit the market in 1980. As the cube grew in popularity it soon became one of the most sold puzzles in the world. At first people's focus on the cube was just to solve it, today more people are interested in how fast they can solve it. The record for the quickest solve is currently 3.47 seconds. The Rubik's cube is also studied in several different fields, for example: computer science, engineering, and mathematics.
Recently I went on the adventure of solving a Rubik's cube myself. I usually don't love puzzles, they make me cry, but this one was not too bad! I did find help along the way from this video by WIRED. Note, there are many ways to solve a Rubik's cube, and my solving was not for time, just for fun. My interest in Rubik's cubes stemmed from my Abstract Algebra class where we have been learning about groups. You may ask, "What is a group? Is that like a gathering of people or things?" Well, kid of, let's define it mathematically;
A Group- To be a group, you must satisfy four conditions:
$1)$ Be closed under the operation. For example let's say we have a group $H$ (for Haylee😉) and let the group operation be multiplication. Let the elements $i,j \in H$. Since the operation is multiplication, then $i*j\in H$ as well.
$2)$ Associativity must exist. Let $i, j, k \in H$, then because of associativity $(i*j)*k = i*(j*k)$, we can regroup our elements.
$3)$ There is an identity element $e\in H$ such that $e*h = h*e = h$ for all $h\in H$.
$4)$ Inverses exist so for any $h \in H$ there exists $h^{-1}\in H$ such that $h*h^{-1} = h^{-1}*h = e$.
*Note that commutativity is not a group requirement, but if the elements in a group are commutative we say the group is Abelian (named after Niels Abel, a contributor to group theory, check out my earlier blog post on him to learn more! ). The group made from Rubik's cube permutations is non-Abelian, so commutativity does not exist.
Now that we know what a group is, the next question we can ask is, what do groups have to do with a classic puzzle? I'm glad you asked!
All the fun of a Rubik's cube comes from messing up each of its perfect faces and then moving the cubies (no, I did not make up this word, it really is what people call the individual cubes on a Rubik's cube) back and forth until each face of the cube is perfect again. It is quite satisfying! In mathematics we would classify these movements as permutations. Here we will describe a permutation as any legal move where you rotate one of the six faces, $\{$F, B, L, R, U, D$\}$, of a $3X3X3$ cube $90^{\circ}$ or one of its multiples. It turns out that the Rubik's cube has $43,252,003,274,489,856,00$ permutations! This means there are $43,252,003,274,489,856,00$ possible arrangements of a Rubik's cube. To see how this is calculated go to page two in this article. These permutations are what make up the Rubik's cube group. But do the movements of a Rubik's cube actually fit the definition of a group?
Before we determine whether a Rubik's cube and it's permutations satisfy the definition of a group, let's look at the make up of the cube and talk about how we refer to each of the pieces.
A cube has $8$ corner cubies, each of them show three faces. It also has $12$ edge cubies, each of those showing two faces. And lastly, there are $6$ center cubies where you only see one face. We will also name the six faces of a Rubik's cube.
Okay, now it's time to make the set of moves of a Rubik's cube into a group!
Let this group be called $R$ for Rubik's cube. Let's show it is closed under the operation of multiplication. Let $M_1$ and $M_2$ be two moves performed on the cube. Since we can make the move $M_1$ and the move $M_2$, then we can also make the move $M_1*M_2$, thus $R$ is closed under multiplication.
Now, let $e$ be the nothing move, meaning $e$ does not change the make up of the cube. Then a move, $M$ times $e$ would give you the same result that you got from just the move $M$, so $M*e = M$. The same idea applies if you did nothing, $(e)$, and then move $M$, you would have $e*M = M$. Thus $M=M*e = e*M = M$, thus $R$ has an identity.
Next let's say that you make a move $M$ and then you reverse that move by performing $M'$. This means you move $M*M'$, first moving $M$ then reversing all the steps of $M$, and the result is as if you did nothing. This would mean that $M*M' = e$, so $M'$ is the inverse of $M$. This idea also works with $M'*M = e$. Therefore, every element in $R$ has an inverse. To picture this better, think about the steps of putting on shoes. First you have bare feet $(e)$, then you put on your socks and then your shoes ($M$). To get back to bare feet $(e)$, you first have to take off your shoes and then your socks $(M')$, and voila, you have bare feet $(e)$ again! So $e = M*M' = M'*M = e$.
Lastly, we must show that the elements in $R$ are associative. Let $M_1, M_2, M_3 \in R$. Let's say that first you move $M_1$ then you move $M_2$ and then you move $M_3$. This would look like $M_1*(M_2*M_3)$ which is clearly the same as if you moved $(M_1*M_2)*M_3$. Since we have $M_1*(M_2*M_3) = (M_1*M_2)*M_3$, we know that $R$ is associative.
And now we know that $R$ is truly a group, woohoo!
So how does knowing that $R$ is a group help solve a Rubik's cube? Well, to solve a Rubik's cube most people follow algorithms, set patterns that you apply in certain situations to move the pieces of the cube. These moves are what we have already labeled as permutations and the permutations are what make up our group $R$. Specifically, these algorithms are made from generators of $R$. This means that mathematically we would let $T$ be a subset of $R$ and we would say that $T$ generates $R$ if $R = \langle T \rangle$ (if $T$ spans $R$, meaning the elements of $T$ can be used to get everything that is in $R$). Here we will let $T= \{D, U, L, R, F, B\}$, our faces from above, so $R = \langle D, U, L, R, F, B\rangle$. Knowing this, you can create your algorithm and you are good to go!
At this point we have successfully touched how the world of Group Theory can relate to a Rubik's cube and seen a little how people can begin to solve a Rubik's cube using mathematics.
Thanks bunches,
~Haylee Jo Lau
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