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.
The devil's algorithm would be the worst, with an average movecount of about 22 quintillion
Correct
easy 1 look
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.)
@@colbytrotter5573 0 look 🤔
@@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.
This one is similar, its solving with only J perms ruclips.net/video/swB17ChOJgA/видео.html
How about the 1S method. The 1 Step solves the cube in one algorithm but requires the knowledge of 43 quadrillion algorithms. Have fun memorizing!
Not within the parameters of my constraints. 1s would average 18 moves, but I was looking for highest move count :)
43 quadrillion? that's just 0.1% of the 43 quintillion you need to memorize.
@@eduardoxenofonte4004 my bad I remembered wrong
@@RowanFortier You can remember 4,3•10¹⁹ algoritms, each contains about Graham's number of movements.
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.
Bogo sort, obviously. Just make random moves.
Old Pochmann, using Nb Perm for both corners and edges.
Under rated channel af
The worst method is moving randomly
Probably, yes. But given the parameters of my restrictions, I think it is mine
theoretically, with good enough luck, actually the best method
Just like bogosort, it is very effective
I would love to see a speedsolve using your method lmao
5:06 do the same thing but use na perms instead of t perm
What about we make random moves until the cube is solved.
That doesn’t match my criteria for the method that I said near the beginning of the video
Got a sub 6 on that method 🗿
I lied 😈
Can’t wait to see you pop off in subs!
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
I remember how young you were when you made a video about a headless chicken
Why?
ruihang xu will still get a sub 1 with the last method 💀💀
minute
wow I can't believe you forgot Petrus
:/
💀
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.
@@MannifyYT2024isn't that x cross
@@squaredude231 not really
Are you going to do a giveaway now that you are sponsored?
I wasn't planning on it. Not sure what I would give away if I were to do one
Wait there are meme algs?
good question! Something like this: ruclips.net/video/ChkqKVTDOJc/видео.html
@@RowanFortier oh wow lol 😄
Devil's algorithm?
beginner is for beginners
No shit Sherlock
You look like Fundy lmao
Lol
hi
Lol
Edit : first comment
Nice pfp
@@justaregularhumanbeing lol