Supercritical Method

[ostracod]

Hello cubers,

A long time ago (maybe a year) I posted information here about a method I made called the Fridrus method. The site is here:

http://web.mac.com/teisenmann/method2/main.html

It became my favorite method, and I could get times under 1 minute. This is good for me, considering I am not focused on speed... If I wanted to go faster, I could memorize more algorithms. If 4 algorithms can get me sub 1 minute, then more algorithms, practice, and finger tricks would make the Fridrus method very fast.

So where is this story going? After a long period of inactivity with the Rubik's cube, I have started experimenting with it again. The result is an exotic mutated Fridrus method, which I call the "Supercritical method". Like the Fridrus method, it uses very few algorithms, relying more on intuitive cube solving skills. The steps are as follows:

1. Solve a 2 by 2 by 2 block.
2. Solve another 2 by 2 by 2 block on the opposite corner, leaving scrambled pieces between the two blocks.
3. Solve any edge between the blocks.
4. Solve corner-edge pairs on either side of the edge from step 3 by breaking things up and using a mini algorithm. There are 3 pairs total.
5. Solve the last layer, which already has a solved corner and 2 solved edges. I've been fiddling with commutators to solve corners...

For more information, go to this page:

http://web.mac.com/teisenmann/Supercritical/main.html

And for a mediocre video:

http://www.youtube.com/watch?v=bWU892phCxE

So, of course, I want to hear opinions. Has this method been made before? Is it a practical method for solving the cube? I haven't practiced it enough yet to know for sure. I am open to suggestions.

-Ostracod

 

[Ellis]

It's an interesting method. Is it practical for solving the cube? Sure why not. It may not be so practical for speedsolving though, it looks like there would be a lot of turns. The fastest/best methods work so well because there is a lot of free space to work with. It seems like after you finish the two 2x2's there's literally no turn you can make without having to temporarily misplace something thats already solved. I don't see the point in restricting yourself. I'm not saying you should toss the idea though, it's cool to make up your own method and make it fast.

 

[DavidWoner]

The two 2x2x2s are not bad after you get used to it, but the place one edge part sucks. I could only figure out 1 way to place edges, and 4 times in one solve the edge I placed was misoriented. I finally gave up and used a 2 flip to solve one of the ones in place. The first two pairs aren't bad, although I mostly just conjugated the 2 blocks into a psuedo F2L so I had an essentially empty layer to work with. The 3rd pair is very difficult. Once you have the third pair you are left with 3 unsolved corners and 2 unsolved edges. Here is the problem with that: Unless the 2 edges are at least properly permuted, you will have created parity and will be unable to solve the corners using a single commutator. I encountered this problem on all 3 of the solves I attempted. I think this method is a little like heise, except severely out of order. Imagine a Fridrich-type method where you solve LL corners after doing the cross, and you have to preserve those corners throughout the solve, and thats how this compares to Heise.

 

[ostracod]

Ha ha... I'm surprised you actually tried it. XD I'm surprised that you had trouble placing an edge, I thought that was the easiest step! And I agree that commutators aren't great for the last layer... Even after fixing edges, I find that commutators don't always work, or at least I can't find a way for them to work.

Here is a turn count from me solving it.

First block: 12
Second block: 9
Edge: already solved
First pair: already solved!!
Second pair: 16 (unlucky, probably makes up for the last one...)
Third pair: 9 (average)
This last layer would take me 3 because I'm so lame. x_X
But let's assume that I learned all of the ~40 algorithms for the last layer. It would probably take me 12 moves.

So, for a somewhat lucky solve done by a "master", it would take 58 moves. I agree, the move count is high. :|

For my own purposes, I'll do a Fridrus count as well:

Block: 20
Edges: 3
First pair: 13
Second pair: 16
Third pair: 4
Last layer: I'll go with 12 again...

68 total. Even if you add some moves to Supercritical to account for the lucky case, Supercritical is still about the same, perhaps better.

Why do I punish myself so? XO

 

[nitrocan]

I AM DEFINITELY GOING TO USE THIS METHOD!!!

Why? Because I got a cube skip after the 4th step at my first try!!! (Wow, what are the chances )

Here is a tip: This would be MUCH faster, if you found algorithms for the 4th, 5th and the 6th steps. There won't be many.
This method has an incredible skipping potential.

 

[ostracod]

Hold on hold on... 6th step? For Fridrus? There is no 6th step in Supercritical. But I'm happy that you see potential in one of my methods!

 

[nitrocan]

Yeah, I didn't really get the 3 pair inserting part in Supercritical. It was way too hard to figure out an easy solution with so many restrictions.

 

[bluephoenix1993]

Can you make a tutorial video about this method in youtube? Well, somehow, I could get step 1-3 because I have encountered blockbuilding methods like Petrus and Roux... They were my first methods before I turned to Fridrich...

Well, in this method, I'm kinda stuck at the 4th step, the one w/c we have to make 3 pairs around the edge between the 2 2x2x2 block...

Uhm.. Once again.. I like this method.. I have been searching for weeks wanting to find a new method, and this one is good...But, can you make a tutorial video? Thanks...

 

[qqwref]

I did a bunch of solves with this method. I think I get around 50-60 seconds with it. The last layer is quite tricky - I was using whatever I knew and managed to get about half of them in one step, but I do think you could realistically learn every case. The main problem I see is the first and second c/e pairs; recognition is quite difficult and sometimes you have to move pieces around a long distance to get them in place. But it's an interesting and very fun method to play with, and it's cool to be able to solve everything but the last 5 pieces completely intuitively. Unfortunately I think it has way too many cube rotations (and movement is a bit too restricted after the second block) to be suitable for speed.

 

[ostracod]

To blue phoenix: I'll make a video just for you.

Wait a few hours until I give a link to a youtube video. I'll try to be as clear as possible about step 4 in Supercritical.

And to qqwref, I agree, it's a fun method, but not the fastest.

 

[ostracod]

Alrighty, here it is (so much for a few hours):

http://www.youtube.com/watch?v=uc54nUzw1EY&feature=channel_page

Good luck!

 

[AvGalen]

Cool method, but useless for speedsolving because it has:
* Soooooo many cube rotations
* Probably the worst look-ahead of any method I have ever seen

 

[ostracod]

AvGalen, I think you summed up this method. I should probably revert back to Fridrus before it's too late! XP

Now someone stop me before I try this: Is it possible to solve the cube with NO method at all? An "improv" method, if you will, where the cuber solves whatever the hell he/she wants until the cube is done? I've been pondering this for a while.

 

[fanwuq]

AvGalen, I think you summed up this method. I should probably revert back to Fridrus before it's too late! XP

Now someone stop me before it's too late: Is it possible to solve the cube with NO method at all? An "improv" method, if you will, where the cuber solves whatever the hell he/she wants until the cube is done? I've been pondering this for a while.

Yes. I do that every time with FMC.
1. Random blocks, probably pseudo.
2. Perhaps finish F2L, perhaps not.
3. Leave only a 3 or 4 cycle or parity to fix.
4. Insert commutator.

I can do this too speed solving and get sub-30. It's call 3LLL petrus.
Intuitive blocks, Niklas to permute corners [U'L'U,R], Sune to orient corners [forgot how to show this as a commutator], U perm (Setup to M'U2MU2, reverse set up Conjugate?).
Of course you can do the same method not intuitively. It's all a matter of perspective and what you understand...

 

[patrickpoako1]

I've tried this method long before you even posted it. Not actually the same but somewhat similar.

The worst things about this kind of method are rotations and look-aheads. Since you're looking for so many pieces at a time you are too restricted to be fast enough. And the intuitive part is almost the ones that is holding as back to become really fast. You see intuitive requires you to think first. In speedsolving you need to limit that or none at all for you to achieve the limit.
Really fast method but somehow not good enough to beat the real thing. ^_^

I started with this method but ended up liking Fridrich and Roux more. Was once a Petrus user but now mainly Fridrich. Sub-25 Fridrich, Sub-40 Roux PB 16 lucky solve. non-lucky 18.

 

[ostracod]

Fanwuq, it sounds like you've gotten a lot of success using your method. I think I should try it as well...

It seems like an improv method is against human nature to make things go in an orderly sequence. I'm going to have to try very hard to solve randomly.

 

[ostracod]

Aghhh, it's difficult! When I try to do it, I always end up with something Fridrusy, Supercritical, or both. It seems like it would take a lot of practice.

 

[Lord Voldemort]

Someone correct me in I'm wrong, but it seems like you can solve the last layer in a relatively doable number of algorithms. First off, the two LL edges are either solved or unsolved (2 Possibilities). If it's solved, then there are either 3 wrong corners (2 Cases) or 2 wrong corners (3 cases). That's 12 orientations. I believe there are 5 possible permutations, so 5 x 12 = 60 LL cases.

 

[ostracod]

Lord "you know who", you're very right, and the small number of cases for Fridrus/Supercritical last layer is a good thing. And if you count mirrors/reversals/conjugates, you need to learn far fewer than 60 algorithms, so learning all the algorithms for the last layer is definitely within reach. It's just a matter of which method can get to the nice last layer the fastest...

 

[qqwref]

Have you guys ever tried 'tripod'? You start with F2L minus a c/e pair (use Heise, Petrus, Fridrich, whatever) and then build a 2x2x1 block on the LL, so you have a sort of tripod of 4 corners and 3 edges left. Then you solve any of the three available c/e pairs, and finish with this LL. I think this might be one of the better ways to reach that nice LL.

Incidentally it is also possible to solve LL by just making a 2x2x1 block (algorithmically) and then doing this. It's an interesting way to do the last layer.