fast(est) cf users

It varies, usually I'll get two R edges in an alg and then get the third while trying to influence the last five in some positive way. Sometimes I can KCLL an edge into place during CLL so I'll solve another two in one alg and move onto L5E.

Then I just do the 'waterman' L5E finish.
 
Sure, sure. This is basically what I do when I do CF. I was just saying that if you're asking "what's the fastest time anyone has with CF" it's not fair to count Roux times, since that method is better than CF methods.

Well for me Corners first is better. I actually average less turns, and I have a better execution of the algorithms, and look ahead. I probably average 46 moves with Roux. I average less than 44 moves with Corners first.
 
... but I always solved 3 redges at a time (intuitively) rather than solving 2-3 ...

Who do you solve 3 redges intuitivly?

(My intuition does not go beyond 2 redges at a time, and does not alway tell me an optimal solution. 3 redges I only get when I setup an easy ELL case sometimes (rare))
 
Who do you solve 3 redges intuitivly?

(My intuition does not go beyond 2 redges at a time, and does not alway tell me an optimal solution. 3 redges I only get when I setup an easy ELL case sometimes (rare))

It really involves an understanding of how you're placing the redges. Basically you can figure them out if you understand how the 2 redge algorithms work. I never really learned all of the 2 case algorithms either. I just looked to see how they worked, and figured out the other cases.

Oh, and by the way, I did find the book where I wrote my averages down. I have a 14.03 avg 10 of 12.
 
i think people are confused with the term CF and the difference between CF and CFOP.
CFOP = cross f2l oll pll. fridrich
CF= corners first method.
CFOP is not CF. if you say anything about your times with CFOP, you are posting on the wrong thread.

On topic, im not anywhere near sub 20 with CF, but i do think CF is a promising method. im sure many dead end thoughts have been put into CF methods, but were starting to talk about CFOP zeroing. im sure there is something you can accomplish with CF methods to make them more speedcube friendly. CF zeroing anyone? i personally was trying to think of ways to do like an ortega for the corners and solving the E slice edges to form the 2 block and finishing with L6E, but i gave up (sounds stupid, but i was trying to see if there was ways to develop nice tricks) i think using the basic CF method purely is kinda poor for speedcubing in comparison to what we have now with CFOP, roux and zz for example.
 
I never really learned all of the 2 case algorithms either. I just looked to see how they worked, and figured out the other cases.

When you start placing 3 redges - I believe you have a complete plan for tow redges do you already exactly know how to solve the third redge or do just know at wich point you have to improvise?

Hope thats clear, have you ever seen a tutorial somewhere.
 
When you start placing 3 redges - I believe you have a complete plan for tow redges do you already exactly know how to solve the third redge or do just know at wich point you have to improvise?

Hope thats clear, have you ever seen a tutorial somewhere.

When I have the last ledge already placed (as in the waterman method), I place 3 redges at a time in one look. I simply don't have the time to make a tutorial, and as far as I know, there aren't any out there for exactly what I do. I can give an example of how this would work.

Use this to generate a case where you'd solve 3 redges at a time:

D2 L' R B2 L F2 U' B F U2 F2 L' B' F' R U2 D2

You should have the corners solved on the right side, and the First layer finished on the left side. You can still get 3 redges solved at a time using URM group of moves.

Here's the sequence I'd use to solve 3 redges for this case:

R2 U2 R U M U' R U2 R2 U2

Now I don't know all the possible cases, but I seem to know enough of them to get by.

So what happens when I haven't placed the last ledge and I want to solve 3 redges at a time? Well I basically try to combine the 2 redge cases on Marc Waterman's stage 3 java page, as well as the sequences in his original booklet (found here: http://rubikscube.info/waterman/booklet.php) to solve 1 redge and the last ledge.

I read what I typed, and I can see that it would be hard to understand, so I guess I'll find another example

First apply this to your cube. I have the white side to my left, and the red side in front of me:

L' B U2 R2 U2 L R2 D' B' R2 F R B D F2

You should have the first layer minus one ledge on the left, and an R' would solve all the corners on the right.

I'm going to split this one up a bit so you can try to follow what I actually did.

The first thing I noticed is that I have 2 edges in the middle ring that are positioned like a familiar 2 redge pattern. The yellow and green at FD, and the red and yellow at BD.

I can use 3 turns to solve the red and yellow edge, and get the second edge into position to be solved. If you look at the second case on the waterman method cage, these 3 turns are the same as the first 3 in this case.

So now I apply the sequence:

(U M2 U')

Now I see that there is a blue and yellow edge in the green and yellow edge spot. I've also noticed the white and blue edge at UL. If I were to perform (R U M U'), I'd have the green edge solved, but I'd be in a bad position to solve one more redge and the last ledge. Instead, I perform this:

(r U' M' U)

Now I'm able to position the last ledge and a third redge using this:

(R2 U M2 U')

Of course if you get good enough at recognition, you will be able to plan that whole solution for the last ledge, and 3 redges using only one look. For most cases, I'm able to do it in 1 to 2 looks.

That whole sequence together is this:

(U M2 U')(r U' M' U)(R2 U M2 U')


After looking at my post, I have realized that there is a bit of a description to solve 3 redges at a time on the Waterman method website.
 
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...Here's the sequence I'd use to solve 3 redges for this case:

R2 U2 R U M U' R U2 R2 U2

...

Thanks for your help it really got me started with 3 Edges at a time. The quoted alg is an especially funny pattern just to use R2U2R2U2R2U2 or its fragments at the end of the sequnence.

I never bothered about planning ahaed for 4 edges at once , but I'll try that too - it's a good trainig anyway.
 
Thanks for your help it really got me started with 3 Edges at a time. The quoted alg is an especially funny pattern just to use R2U2R2U2R2U2 or its fragments at the end of the sequnence.

I never bothered about planning ahaed for 4 edges at once , but I'll try that too - it's a good trainig anyway.

It does take a while to get used to it, but It ends up being better than solving one edge, then searching for the next edge, etc.

That's just one of the techniques you can use to solve the redges.

Ah, that makes sense, first I was a bit confused as the M move seemed to insert FR in the wrong place.

I'm glad it makes sense now.
 
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