He got it wrong. What he calls God's Algorithm is actually Devil's Algorithm. A God's Algorithm is a specific algorithm that solves a given scramble in the fewest moves possible (God's Number or less)
@@SB-ex2px Cant you just concatenate a bunch of blind solve methods and their reverse to get gods algorithm? Like say you manage to order all 43 quintillion combinations in a way that makes sense, gods algorithm would just be the blind method solve of the 1st scramble, and then the reverse, then the blind method solve of the second scramble, then the reverse and so on, 43 quintillion times. I bring up the blind method bc if you know which edges and corners have to go where it seems way more deterministic than the friedrich method, where you dont know what OLL and PLL you have at the end, and there are way more "choices" so to speak. In the blind method you just have to choose which new cycle you're gonna go for, but if you just label each edge and corner and say you have to go alphabetically when choosing a new cycle, then theres no more choice involved. This algorithm would have maybe 17.2 sextillion moved so its pretty ineficient, but I feel like its not that hard to construct
@@SB-ex2px Oh so is gods method a way to figure out the 20 move algorithm for every scramble? Cant you do that computationally relatively quickly? What are you working on exactly then?
Does it take a lot longer for a computer to find the shortest way to solve a cube rather than just simply solving it? How quickly does, say, the latest iPhone or Samsung Galaxy to solve a cube? Seconds/ milliseconds/ microseconds/ nanoseconds? I doubt it's femtoseconds yet...
Any possible position can be solved with 20 or fewer moves!! It so find optimal solution(solution that takes least moves) Let's start from solved state; We have 18 possibility moves at solved state (r r' r2) same for 6 other sides. For second move we have 15( as, last move will not consider) Then all throught to 20 moves we have 15 combination from last move. So we get, Combination-moves 18. 1 move 18×15. 2 moves 18×15×15 3 moves To the.......20 moves. So each time we do a single move we will get new state of cube. So by combining all the other 15 moves to last state to the 20 moves we can find the optimal solution at once. Well i started from solved state. But we don't have to do that... Starting position: We can do from any scramble. So doing this combinations we can get the solved state at once!! So if u understand my thing then see my question: Question: I think at each combination we can new state( e.g. R , R U , R U F, R U F B') So is that possible that some time doing combination can we get a state that is already happened before ( e.g. If i do R U F' B U' L R Now getting this state can also be get by e.g. U' F L D B' //DON'T TRY THIS MOVES I'M JUST TAKING EXAMPLE) So is this possible?
Interesting question! I'm not sure I entirely understand what you're saying in the first part, but I think I get the gist. It is certainly possible to arrive at one permutation from different ways. For example, there are many different solutions to the Rubik's cube, meaning there are many different ways to arrive at one permutation. That's the easiest and most obvious example I can think of: 1 permutation, many many different ways of solving. Certainly possible. Now, the significance of this question comes into play when thinking of God's Algorithm. Consider my second definition of God's Algorithm: one algorithm that can simply and shortly solve the Rubik's cube, without being repeated a necessarily integer number of times. So, I defined the shortest length of this algorithm by the number of permutations divided by the maximum order of any algorithm, resulting in about 34 quadrillion if I recall (I'm not gonna bother getting the exact number). Remember, this is the SHORTEST length of a "God's Algorithm" that fits the second definition. However, realistically (it's difficult to say "realistically" when we're talking about algorithms of quadrillions of moves) this algorithm could be complicated by several factors. First, to be only 34 quadrillion moves long it would have to have an order of 1260, the longest possible value for an order. However, it is possible to have an order much shorter, so that is one complication. The other complication is the one you have brought up: what if some of the permutations are covered again in the algorithm? That means a 34 quadrillion length algorithm would not be covering 34 quadrillion permutations, thus the algorithm would have to be longer to cover all 43 quintillion possible permutations. So, hypothetically, in creating such an algorithm one would have to ensure that each move covers a new permutation, never covering one which has already been covered in the algorithm (or in previous repetitions of the algorithm). However, I would dismiss that definition and its technicalities and complications as impractical. Such an algorithm will likely never be found, and unless you have a whole lot of extra time on your hands I wouldn't recommend trying :)
There’s 3 ways to make the cube impossible: corner twists, edge flips, and edge swaps. For corner twists, a multiple of 3 means the cube is fine (1 in 3). For edge flips, an even number means the cube is possible (1 in 2). Same goes with edge swaps: even number is good (1 in 2). If you were to take the cube apart and assemble it randomly, there would be a 1 in 3 times 1 in 2 times 1 in 2 chance of it being solvable. That comes out to 1/12 (8.333... percent) chance. Each possibility is 1 of the 12 orbits.
So I watched this video until 19:59 and now you're telling me that the whole reason I clicked on this thumbnail to watch this video does not exist. Soo annoying. Well at least I can solve the cube using the Friedrich Method, Roux method and APB method method.
43:40, I found the super Flip not solved by twenty moves, it's can solve by 19 moves by middle part turn: U' R2 U' F2 D' R2 U' B F. U2 L' B2 R2 F2 (E!) R' U2 F' B' Note: (E!) Replace two unoptimized moves which are; D' U
Very good, I loved this video. I agree with the numerator of your equation (12 !) x (8 !) x (2^12) x (3^8) but I don't think 12 is good enough for the denominator of the equation. You asked your audience what was wrong with the equation and someone correctly mentioned that it fails to address the fact that each of the 12 corners can be placed in a unsolvable ("illegal" you called it) position. Then you put 12 as the denominator in your equation. I suggest that each of those 12 corners can be positioned in 2 different orientations to create an unsolvable state. I argue that 24 is a better number than 12. Furthermore, no one mentioned the illegal edge situations. There are 12 of those as well. You astutely recognized that flipping 2 edge pieces takes us from even to odd, so two "wrongs" make a "right". I think that half of those illegal edge swaps = 12/2 = 6. Maybe the correct denominator is 12 x 2 x 6 = 144. Performing the math gives us [ (12 !) x (8 !) x (2^12) x (3^8) ] / [ 12 x 2 x 6 ] giving us 3,604,333,606,207,488,000. Also, speaking for the centers which some mathematicians say "they don't move". You acknowledged (correctly so) that they do in fact move ... they rotate. You go on to say that "it doesn't matter" which is partially correct. For "COLOR ONLY" it does in fact not matter, but for "ORIENTATION" it does matter. If you were to take a magic marker and draw arrows that all face upwards on each piece of the cube, you will find additional mathematical permutations that do exist. It shows up on the 6 center pieces and can be in one of 4 possible orientations. The arrow can point up, right, down, and left. The math now begs for those possibilities, even if the color of the center does not change.
12 is the correct denominator The orientations of the 8 corners (which have 3 orientations each) must sum to zero so multiply by 1/3 The orientations of the 12 edges (which have 2 orientations each) must sum to zero so multiply by 1/2 The permutation parity of the whole cube must be even since every move keeps permutation parity even so multiply by 1/2 1/3 x 1/2 x 1/2 = 1/12
@@fatimazahrae7654 Basically, the orientation of the last corner (or edge) is dependent of the orientation of all other corners (or edges). So divide by 3 for corners and 2 for edges (the number of orientations they have), then finally divide by 2 since permutation parity is always even
@@charlieharrison This algorithm is instructive right here: R U R' F' R U R' U' R' F R2 U' R'. Whenever two adjacent corners are swapped, you'll find that there also must be two edges that are swapped. There is no way to put the cube in a position where only two edges are swapped without two corners being swapped as well. After you swap those corners with the above algorithm, you'll find it's impossible to keep those corners swapped and solve all the edge pieces.
This is mentioned in the wikipedia article on "Rubik's Cube Group". But either there is an error in the algorithm (R U2 D' B D') or they didn't test it correctly. The mentioned algorithm loops already after 36 repeats, NOT 1260. I have found a simple alg (L R' D U B) which loops after 3960 repeats. See my video here: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-Xs27H_sdM1E.html
The way he explained parity is wrong and incomplete. The two edges (they are called wings by the way) are swapped with each other (you cannot flip a wing). If you swap them, it looks like the big edge is flipped but you literally cannot flip wings (try taking one out in a 4x4 and putting it back flipped). By the way, parity has to do with how many slice moves you did, so if you did an odd number of slice moves since the solved state, you will get parity.
You can call them "wings" if you want but they are in fact mathematically called "edges". You can definitely "flip an edge" in every sense of the word. The terms "slice" and "parity" typically refer to an [even#] by [even#] cube, for example a 4x4x4. This author is talking about a 3x3x3.
rainbow tyrant I found that out after . I just kept coming across that video . So I tried it. It labeled it as a fake video in the description (which I never read) 😂🤷🏻♂️
@@ronniebrummett4995 noice i actually did the same thing witha different video when i started...if you want to learn the best and easiest to under stand is the tutorials from rubiks channel.
I've seen quite a few of those videos! Nathan Wilson's famous one actually inspired me to think about this topic more, and clarify that such an algorithm is not possible. I would definitely recommend taking some time to try learning it! Maybe even try without a tutorial for a while first.
I prefer to use the minus sign for anticlockwise rotation. Let's name each face of the cube as Right, Left, Up, Down, Front and Back. Then define each rotation as a 90-degree clockwise rotation. R: One 90-degree clockwise rotation of the right face of the cube. 2R = R+R: Two 90-degree clockwise rotation which is equal to 180-degree clockwise rotation. -R: One 90-degree anti-clockwise rotation. 4R = 0: Four rotations is equal to zero since it is a 360-degree rotation and the cube doesn't change. R = -3R: One clockwise rotation is equal to 3 anti-clockwise rotations. With this notation it is possible to add and subtract the rotations. The only problem is that the addition is not commutative: R + U ≠ U + R
There's plenty of ways that a Rubik's cube is related to math! One of the biggest connections between math and the Rubik's cube is group theory, but there's plenty of other connections too.
If I explain how to find the exact number of total combinations that can occur and thus prove that it is almost impossible to find a solution by random moves, would it be a good idea? How does it relate to mathematics again?
