Beginner’s Rubik’s Cube!

I’ve brought one Rubki’s cube to college from home in freshmen year. Now I’m an old senior, I still haven’t been able to solve these cubes. Plus, I’ve got three more from various companies at career fairs. Finally decided to learn about it.

I was able to do the first layer, but always get stuck on the second.. So I found a beginner’s video. The algorithm presented here uses the same logic for the second layer as for the first layer–try to interchange two pieces without interrupting the desired layers. Solving the third layer, however, is much trickier and involves a lot memorization.

As I’m feeling more comfortable using this algorithm, I started to ask questions.

—————-How many step do we need for the first two layers?——————-

I tried many times, and counted the number of steps I took for the first layer. They range from 21 to 34, but usually around 26. The second layer usually take around 4*12=48 steps, but in the worst situation we might need to even double the number. From wikipedia, the optimal algorithm can solve the entire cube in 26 steps. So this algorithm, though intuitive, is far from optimal.

———————If we fix the bottom left 2*2*3 blocks, and restrict our operations to Uc and Rc (upper clockwise and right clockwise), how many configurations can we have? Also, what is the order of UcRc? In other words, if we are further restricted to Uc immediately followed by Rc, how many steps do we need to get back to the original configuration? (It’s certain that we are able to get back to the original configuration. There are only finitely many configurations. So we must have (UcRc)^i=(UcRc)^j. Then (UcRc)^{j-i}=id.)————————————-

In this setting, only 13 blocks are actually involved. The other 13 remain fixed rotations in this restricted group. We know that corner pieces only go to corner pieces, and middle pieces only go to middle pieces. Suppose these are the only constraints, then we have 3^6*6! possibilities for the 6 corners involved, and 2^7*7! possibilities for the 6 middle pieces involved. That’s a total of 3^6*2^7*7!*6!=338610585600 possibilities! Intuitively, I don’t think all are attainable, but I don’t have a good reasoning for now.

Nevertheless, I’ve already learned something from this to be answered question. By writing down my reasoning why we can always go back to the original configuration in this restricted setting, it came to me that the same reasoning applies to any other restricted or non restricted settings. The group (easy to check it’s a group!) of operations on 3*3*3 rubik’s cube is generated by Uc, Dc, Rc, Lc, Fc, Bc (with certain relations yet to be figured out). Each of these generators have order 4. Other elements in the group might have very large order, but never infinite–not because the generators all have finite order (think about free group), but because the object we act on has finite configuration space.

 

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