Bindedsa
Veteran
why would you do this
perhaps it's easy to find F2B edges? I can't image it takes more than 3/4 moves extra most of the time.
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why would you do this
perhaps it's easy to find F2B edges? I can't image it takes more than 3/4 moves extra most of the time.
Edit: I'd like to see some fast CFOPers try and solve F2B using this method, and see if they can get anywhere close to Alex Lau's F2B times (I doubt it, but as I have said, it's a pretty fun method to play around with anyway).
it's basically just f2l. i think with little practice, most cfoppers can get this equal or slightly faster than their regular f2l times.
I think it would take a decent amount of practise - the possibility of an F2L edge being in the D layer would hinder lookahead quite a bit, I think...it's basically just f2l. i think with little practice, most cfoppers can get this equal or slightly faster than their regular f2l times.
I think it would take a decent amount of practise - the possibility of an F2L edge being in the D layer would hinder lookahead quite a bit, I think...
Ok, that's a good enough reason for meIt's actually pretty fun.
Wojciech Knott got a 6.56 clock average with a counting 5.39 and 7.72... ><
http://cubecomps.com/live.php?cid=831&cat=7&rnd=1
I think it would take a decent amount of practise - the possibility of an F2L edge being in the D layer would hinder lookahead quite a bit, I think...
Ok, that's a good enough reason for me [emoji14]
I would find it hard to put the pairs in the right slot when I don't have centres there unless I don't rotate - unless you solve the centres too, in which case you're basically doing CFCE?
I might not have clear enough in my description of the method, but I ensure that there are no F2L edges in the D layer while making the cross.
that's not what he means. i think what he is saying is place two opposite cross edges and put any bottom or top edge in any orientation into the remaining cross slots. then solve 4 f2l pairs.
lol on cam
Generated By csTimer on 2015-3-1
avg of 5: 7.12
Time List:
1. 7.63 R2 U2 B' D2 F U2 B U2 L2 B' D2 R' B' D R D U2 B F2 L2
2. (5.15) L R2 B R2 F L2 B U2 B2 R' U' L' R U' B' F' D U'
3. (8.42) B2 R2 B D2 L2 F2 D2 F2 D2 R2 B' L U2 B D R D2 L' F D' U
4. 7.94 R2 D R2 B2 R2 B2 U' B2 R2 D' L2 B' D2 F' U R D2 B F L U2
5. 5.79 D U B2 F2 U' B2 D' L2 U2 L2 B R2 D' L F2 R' D2 L2 B
7.94 could be easily faster, I started doing a J perm from the wrong angle![]()
Yes, but when you solve the not-cross edges at the start, do you care about how the M slice is oriented, or not? I was assuming not, but I think you're solving it too...
I know it doesn't make it CFCE, but it's very similar, and I think that if you're going to 'solve' the two edges in the D layer then it's not much different to just solve a cross and do CFCE; it's probably the same speed, if not a bit faster.I solve the centres as well because as you said, not having them solved would make recognition harder. I haven't really tried it though. I suppose it would be faster.
I don't think solving the centres would make it CFCE, because you're not solving the DF and DB edges until the end. CFCE does not have a LSE step.
I solve the centres as well because as you said, not having them solved would make recognition harder. I haven't really tried it though. I suppose it would be faster.
I don't think solving the centres would make it CFCE, because you're not solving the DF and DB edges until the end. CFCE does not have a LSE step.