Solution of Seven-Order Rubik's Cube

After the broadcast of My Little Confessions, there were many negative comments on me online. To sum up, there are three main reasons for supporting the contestants: first, he is a master of folk science, and I myself have become a "senior official" in science. The truth of the matter is: Chen Daji is a graduate of the Central Party School, a 100% China civil servant, and a typical bureaucrat; Our association is a spontaneous and loose folk Rubik's Cube research society. I have a false name as the vice chairman of the association. I don't have a penny of salary, but I still have time and money. I am a diaosi who works hard in Qian Qian. Second: he is the world record holder, which is his aura of boasting and fooling people. The fact is: there is no Rubik's cube or Rubik's cube of the highest order in the world, which is simply an international joke. The following is the perfect Rubik's Cube formula of prime number P greater than 7: a (i, J, K) = (I+2J+4K) * P+(2I+4K+J) * P+(4I+J+2K) mod (P). The so-called highest order-especially 15 1 order-is just an ignorant and fearless international joke.

The initial position chosen by David is A (a, B, c)= (6, 6, 3) after the player's perspective adjustment, that is, the starting number 1 is placed in the sixth row and the third column on the sixth floor. The parameters in this seventh-order standard Rubik's Cube are: 1, and the initial position A(k0, i0, j0)= A (a- 1, b- 1, c-1) = (5,5,2) =/kloc-. 3. The small steering amount (k2, i2, j2)=( 1, -2, 1), that is, when it encounters a multiple of 7, instead of a multiple of 49, it becomes 4, and the large steering amount (k3, i3, j3)=(-3, 4, -2). I'll tell you one thing, the big steering vector doesn't need to be memorized-the player may not know this, because he basically converts it through exchange and then memorizes it, and he doesn't understand the profound principles and laws behind it at all. Otherwise, if he knew, he wouldn't have spent months tossing about this matter, and he wouldn't be so arrogant that he was the only one in the world who could know. All these signs show that he doesn't know the profound principle of this method, but only knows why. I have demonstrated that if the initial position is A(k0, i0, j0) (this is in the system represented by (0, P- 1)), then when the large steering vector is (k3, i3, j3)= (2*k0+ 1, 2 * i0+. For example, since A(k0, i0, j0)= (5, 5, 2), there must be (k3, i3, j3)= (2*5+ 1, 2*5+ 1, 2 * 2+6558. 5) mod (7), please note that (4,4,5) mod (7) and (-3,4,2) mod (7) are completely equivalent in actual construction effect. The walking method is exactly the same as the plane continuous pendulum method known in 17 and 18 centuries. In order to take care of the general public, let's give a few examples to illustrate. In the system represented by (0, P- 1), because A(k0, i0, j0)= A (a- 1, b- 1, c- 1) = (5, 5, 2) = According to the continuous vector (k 1, i 1, j 1)=(0,-1, 3), the next position is A (5 5,5,2)+(0,-1, 3). According to the continuous vector, the next position is a (5,4,5)+(0,-1, 3) = a (5,3,8) = a (5,3, 1) = 3, that is, the number 3 is placed in the second column of the fourth row on the sixth floor; By analogy, A(5, 6, 6)= 7, that is, the number 7 is placed in the seventh column of the seventh row on the sixth floor. At this time, the first multiple of 7 appears, but it is not a multiple of 49, which needs to be handled according to the small steering vector. According to the small steering vector (k2, i2, j2)=( 1, -2, 1), the next position is a (5,6,6)+(1,-2,1) = a (6,4). Then, as a starting point, continue to play numbers according to continuous vectors (k 1, i 1, j 1)=(0,-1, 3). And so on until a (4 4,0, 1) = 49, that is, the number 49 is placed in the second column of row 1 on the fifth floor. At this time, the first multiple of 49 appears, which needs to be handled according to the large steering vector. According to the large steering vector (k3, i3, JBOY3) = (-3,4,2), the next position is a (4,0, 1)+(-3,4,2) = a (1,4,1). According to the above process, you can fill in the seven-order standard Rubik's Cube one by one by repeating it. The core of the above process is four groups of vector parameters: 1, initial position A(k0, i0, j0)= A (a- 1, b- 1, c-1) = (5,5,2) = 65438. 3. The small steering amount (k2, i2, j2)=( 1, -2, 1), that is, when it encounters a multiple of 7, instead of a multiple of 49, it becomes 4, and the large steering amount (k3, i3, j3)=(-3, 4, -2). This is also why we have the belief that we can teach primary school students PK players surnamed Chen. The reason is that simple. For each position -342 positions, we only need to remember 4 sets of parameters. According to the principle of central symmetry, we only need to remember three groups of parameters, and the large steering vector can be given directly according to the initial position. Theoretically, the total memory is 342*3*3=3078. In addition, if you consider the rotation and reflection angle of the perspective of the Rubik's cube, you don't really need to remember so much, that is, you don't consider any conservation transformation, but only the transformation of the perspective, then the theoretical total memory is 3078/24= 128.25, that is, it doesn't exceed 130 to remember, because each group has 9 numbers, then1. According to Mr. Wen Li's calculation, regardless of any conservation transformation, the actual total * * * needs to remember the parameters of 19, that is, 19*9= 17 1 number can be converted from perspective.