if instead of 1 move you would do 3 jperms (U = x (R2 F R F' R U2 r' U r U2) x' y x (R2 F R F' R U2 r' U r U2) x' y x (R2 F R F' R U2 r' U r U2)), the average movecount would be about 660 quintillion (if the Jperms were done optimaly. if they were done using the normal algorithm most people use, the average movecount would be about 924 quintillion.)
@@comp.lex4 a 0 look would be if you could close your eyes during the whole solve, so you couldn't even inspect, but the devil's algorithm needs to be done multiple times during a solve, and you would constantly need to look at the cube to see when it is solved, and you dont know when to stop if you can't look at the cube
Idea: OLL 40 method, where the only alg is R' F R U R' U' F' U R. Step 1: Use it to permute all the pieces. This is intuitive because it just does a 2 corner swap and a 2 edge swap permutation-wise. Then, orient the pieces using repetitions of this alg. 2 times twists 3 corners and flips 2 edges, 4 times twists 3 corners, and 6 times flips 2 edges. This is also intuitive once you know what the alg does when repeated twice.
CFSEPO Step 1: Cross (12•11•10•9•2⁴ = 190079 algoritms + 1 solved) Step 2: First layer (8•7•6•5•3⁴ = 136079 algoritms + 1 solved) Step 3: Second layer (8•7•6•5•2⁴ = 26879 algoritms + 1 solved) Step 4: Edges on last layer (4!•2⁴/2 = 191 algoritms + 1 solved) Step 5: Permutation of corners on last layer (4!/2 = 11 algoritms + 1 solved) Step 6: Orientation of corners on last layer (i counted 21 algoritms + 1 solved) Edited after 4 minutes: M M' M2 S S' S2 E E' E2 x x' x2 y y' y2 z z' z2 exists.
hear out i made my own method that really efficent.Yeah i know the title of the video but i want to tell my idea to everyone.first you do cross,2nd layer edges, orient the last layer edges, f2l, with a lastlayer cross skip. You can do ZBLL or anything
Sometimes I see I can build a 2x2 block at the start of my solve (sometimes even a 2x2x3 block) so I just do that and then carry on with CFOP normally.
