i need some help with a new method

rocker smurf

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Joined
Dec 4, 2009
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i have been solving rubiks cubes for a while and decided to try to make my own way to solve it, and i did.
i would like to know if there is anyone who is intersted in helping me perfect it.
heres how i did it





1. solve all edges, making corsses on all sides.
2. then find cornors that are not on the right sides ex(yellow cornors on white side) they may not be oriented or permutated corectly. there will always be the same number of 'bad' cornors on both sides. put ur top edge to ur left.line up two bad corners and aply algarithem
L D2 L' U F' D2 F U' F' D2 F U L D2 L' U' and do this to all bad cornors
3. orient and permute cornors of last two layers


for algarithums for first and last step email me at [email protected]
 
I would suggest doing it the other way around, corners and then edges. I got down to 50 seconds with CF in 3 days.
 
It doesn't sound like it has much potential. That algorithm could take a while. Although you could OBL & PBL for the last part, but that would be around 60 algorithms.
 
i got to a mitute thirty today and i just started it today. you use the same algarithem for the last stepbut you apply it difrently and there are only three for the first step so its easy to learn
 
i got to a mitute thirty today and i just started it today. you use the same algarithem for the last stepbut you apply it difrently and there are only three for the first step so its easy to learn
I meant if you wanted to expand it, it would be faster to use OBL & PBL. Right now you'll probably max out at 50 seconds.
 
k thx
if u want i would apreciate anyone would want to help work on it so just ask for the rest of the algarithams
 
Your methods pretty good, i tried it.
I got 40 secs. Its better than beginners,
but i think that fridric is da best
 
For step 2. I would prefer something like (R' F R F') * 3 or (R U R' U') *3

About step 3:
How many substeps do you need to do this?
I take it there are alot of algorithms to do it in one step.
2 steps should be fairly possible, but still sounds pretty ambitious.
 
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L D2 L' U F' D2 F U' F' D2 F U L D2 L' U'

x R2 U' L' U R2 U' L U x'

Or, since you haven't yet oriented corners, you can do the same permutation of corners using:
U L D2 L' U' L D2 L'

This moves the corners in the same way as for as position is concerned, seeing as how you are going to orient later anyway.

Chris
 
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