This is a batch of reference materials for you. Have a nice day!
Although the Rubik's Cube has only 26 small squares, it changes a lot. The total number of changes in the Rubik's Cube is 8! *3^8* 12! * 212/(3 * 2 * 2) = 43252003274489856000 species or about 4.3 10 19. If you can turn the Rubik's Cube three times in one second, you will need to turn it for 454.2 billion years, which is about 30 times the current estimated age of the universe.
The reason for the total number of changes in the third-order Rubik's cube is this: after the six central squares are oriented, a coordinate system is formed. In this coordinate system, all 8 character blocks are arranged in 8! And each character block has three orientations, so it is 8! * 38, 12 prism blocks are all arranged, each with two orientations, that is, 12! *2 12, so multiplication is a numerator. The meaning of 3*2*2 on the denominator is to keep other color blocks still, and it is not allowed to change the direction of a character block (corresponding to 3), change the direction of a prism color block (corresponding to 2) or exchange the positions of a pair of prism color blocks or a pair of character blocks (corresponding to another 2) separately. As for why, I suggest you think about it first. I wrote some preliminary works. You can go here and have a look.
It can be seen that it is not easy to change back to the same color in a short time with so many changes. However, the fastest person in the world can restore a Rubik's cube in 7.08 seconds (the record was set at the Czech Open on July 12, 2008), and the record holder is Erik Akkersdijk from the Netherlands.
Why are those people so fast? Because he can remember many Rubik's cube algorithms, or there are also top players in the world called Rubik's cube formulas. It is said that he can remember more than 600 algorithms. The introductory Rubik's Cube solution we introduce here involves few algorithms and is very simple. As long as you learn, everyone can easily learn to play the Rubik's Cube.
Before we begin, let's look at the basic structure of the Rubik's cube. The relative positions of the six central squares of the Rubik's Cube are fixed. You'll know this when you dismantle the Rubik's cube. I'm sure that when you start wriggling in the following way, it's easy to forget what color you started with, so it's messy. So be sure to set a direction you like at the beginning. Here I choose blue as the top, green as the bottom, red as the front, orange as the back, white as the left and yellow as the right. Of course, your sticker may be different from mine. There should be 5*3 stickers on six sides of the Rubik's Cube! =30 kinds of stickers, why? Because if you specify the blue face as the top face, then the bottom face should have five options, and the remaining four faces form a ring, except for rotational symmetry, there are three! Is the planting method right:)
I'm going to preach again If you are in a hurry to see the algorithm, you can skip it first.
First of all, we observe two cases: 1 2. In both cases, the three dislocated blocks are in the same position in the sense of rotation, right? This is one of their characteristics. Green has three directions at each corner. If we mark four corners, we might as well call it the case of 1 (from the upper right corner113), the case of 2 (2223), the 3 in brackets indicates that the green block is already on it, and the 3 (1233) is. 6 situation (12 12), 7 situation (2 1 12), have you found any rules? The sum of the numbers in brackets must be an integer multiple of 3! Why does it have to be an integer multiple of 3? I suggest you look at the total variables of the original Rubik's cube and prove that the sum of the angles of the corner blocks should be an integer multiple of 360 degrees. This restriction determines that we can only have eight situations.
This problem is not particularly simple. First, let's ask a question. If the four corner positions on the top of the Rubik's Cube can be marked as 1, 2, 3 as above, they can be marked as (1113) (1233) and so on. The same situation can be removed after rotation.
If the same rotation is not removed, then the four angles can be distinguished. The answer is simply 3 4 = 8 1? But if the same rotation is removed, it will be more complicated, for example,113,311.
This is a classic problem of combinatorial mathematics, called the necklace problem (that is, wearing a necklace with several colors of beads), or called Polya theorem. You can go here if you are interested, but I still suggest you think about it yourself first. The formula here will make you dizzy at once, if you are unfamiliar with the concept of "Euler number" or have never heard of it.
In this case, we don't need the formula in mathworld, just enumeration:) But if we think about it again, the answer is 24. I looked it up and gave the same answer as the horror formula. Among these 24 kinds, the remainder of the "sum" of 4 numbers divided by 3 should be evenly distributed to 0, 1 2, which I haven't proved yet, hehe, so there should be 8 divisible by 3. That's right+these 7 kinds: