# Supercritical Method



## ostracod (Dec 21, 2008)

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


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## Ellis (Dec 21, 2008)

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.


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## DavidWoner (Dec 21, 2008)

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.


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## ostracod (Dec 21, 2008)

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


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## nitrocan (Dec 21, 2008)

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.


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## ostracod (Dec 21, 2008)

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!


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## nitrocan (Dec 21, 2008)

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.


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## bluephoenix1993 (Dec 22, 2008)

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...


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## qqwref (Dec 22, 2008)

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.


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## ostracod (Dec 22, 2008)

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.


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## ostracod (Dec 22, 2008)

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

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

Good luck!


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## AvGalen (Dec 22, 2008)

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


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## ostracod (Dec 22, 2008)

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.


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## fanwuq (Dec 22, 2008)

ostracod said:


> 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...


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## patrickpoako1 (Dec 22, 2008)

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.


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## ostracod (Dec 22, 2008)

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.


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## ostracod (Dec 22, 2008)

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.


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## Lord Voldemort (Dec 22, 2008)

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.


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## ostracod (Dec 22, 2008)

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...


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## qqwref (Dec 23, 2008)

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.


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## Lucas Garron (Dec 23, 2008)

ostracod said:


> Lord "you know who", you're very right, and the small number of cases for Fridrus/Supercritical last layer is a good thing.


No, he's not. His math looks like he never took a single math class at Hogwarts.

It's 2*3^2*6 = 108.
Mirrors don't even halve that. Still doable, but it's always surprising how many cases 2 edges+3 corners have (and about as surprising that everybody assumes it's much less).


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## patrickpoako1 (Dec 23, 2008)

qqwref said:


> 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.



If you look at this on the correct angle its sort of an OLL. I might be wrong or might have not understood you correctly. 
I tried doing this (as from what I understand) and ended up with either sunes, U cases or T cases.

I might be doing something wrong. Can you explain this further?


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## qqwref (Dec 23, 2008)

You're doing it wrong. You're supposed to do the LL in one step, not go "oh yay easy OLL". The point is that there's a 2x2x1 block solved in the last layer; doing OLL will usually break the block up. It is definitely doable to learn all of the one-step LL algorithms for the situations with a solved block.


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## ostracod (Dec 25, 2008)

Qqref,

The Tripod F2L is similar Fridrus, just in a slightly different order... 2 by 2 by 3 block, then 2 edges of the "cross", then 2 corner edge pairs, and you are left with a tripod, and after solving another corner edge pair you get the nice last layer.

And to Lucas, even if there are 108 cases and some number less algorithms to do a "nice" last layer, it's much better than the 1000+ algorithms needed to do a "normal" last layer. I've found conjugates to be a very useful way to cut down on learning algorithms. For instance, if there are two miss-oriented corners diagonally opposite from each other, and I only know how to solve them when they are next to each other, I just move one face to put them together, use the algorithm, then undo the face turn. In fact, I can do a 2 look "nice" last layer if I know 6 algorithms!!! Here's a list:

-Orient edges
-Swap edges and front corners (I use this to solve permutation of both corners/edges. Can also orient edges.)
-Orient edges + cycle corners (I plan to learn this one to solve last layer in 2 steps.)
-Corner cycle
-2 corner orient
-3 corner orient

This is why I use the Fridrus method. I like to learn a small number of algorithms and focus on intuitive solving. ;D

And yes, the point of the last layer is NOT to have a nice "OLL". If you focus on orienting the last layer, you screw up the permutation of the already solved pieces. That's why I solve edges first, then corners.


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## patrickpoako1 (Dec 25, 2008)

ostracod said:


> Qqref,
> 
> The Tripod F2L is similar Fridrus, just in a slightly different order... 2 by 2 by 3 block, then 2 edges of the "cross", then 2 corner edge pairs, and you are left with a tripod, and after solving another corner edge pair you get the nice last layer.
> 
> ...



when you do your 2x2x3 block you only have 1 edge for the "cross" missing, right? where do you put your block btw after you solve it? lower-left part?


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## qqwref (Dec 25, 2008)

ostracod said:


> The Tripod F2L is similar Fridrus, just in a slightly different order... 2 by 2 by 3 block, then 2 edges of the "cross", then 2 corner edge pairs, and you are left with a tripod, and after solving another corner edge pair you get the nice last layer.



Maybe, but my biggest problem with the Fridrus method mentioned in this topic is the 2 edges of the "cross", which just seem to make it more difficult to move pieces around without making the method more efficient. So for that reason I prefer just pure blockbuilding.


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## ostracod (Dec 25, 2008)

To Patrick:

There are 2 edges to solve in the cross, not one; the block stays in the back laying vertically. As I solve corner edge pairs, I break the block up (a lot!).

And to Qqwref:

You do have a good point. I may consider using block building instead of placing 2 edges, but then I won't be able to say I use the FRIDRus method. XP That's OK though.

BTW, where did you find out about the Tripod method, and is there a site where I can get more info? It seems like the method I've been searching for.


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## qqwref (Dec 25, 2008)

ostracod said:


> BTW, where did you find out about the Tripod method, and is there a site where I can get more info? It seems like the method I've been searching for.



I didn't learn it from anywhere, just made it up. It was the end result of my attempts to try to solve as much of the cube as possible with just blockbuilding and no algorithms of any kind.

I tried to find a way to consistently solve the last 7 pieces (the tripod) in two algorithms, but I could never find one. Even with the corner/edge pair approach, there are still many times when it's not at all obvious what to do... but it's closer than ever to being speedsolve-worthy


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## ostracod (Dec 26, 2008)

Qqwref, because I've been messing around with supercritical so much, I am very good at the last corner/edge pair in this method. I can consistently form a pair in a maximum of 8 moves. You can see my site for all the cases. I just need to practice block building before the tripod... I'll probably do some analysis to find the most efficient ways to do so. With any luck, I might be able to get good solve times! X3

Thanks a lot again for all the help.


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## ImNOTnoob (Dec 26, 2008)

Cool! I just got a J perm for the last step!


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## yurivish (Dec 26, 2008)

How many algorithms would one need to solve the cube after a "tripod" with all edges oriented? If you do a ZZ start and then solve the 3 slots and top part, that's what you end up with.


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## ostracod (Dec 26, 2008)

Yurivich, I don't think that orienting edges before the tripod would be very helpful, because by solving a corner edge pair in the tripod, you would ruin the edge orientation. It would help, however, if there were specialized algorithms for such a situation.

I've actually been on vacation in Florida for a few days, writing from my iPhone. When I return to my computer, I'll be able to do more work on this method and search for more algorithms. Perhaps I'll make a new site for the tripod method...


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## ostracod (Dec 26, 2008)

Stuck in lots of traffic... Only have iPhone + rubik's cube... So time for a forum post!

I think that the tripod method is one of the most versatile methods, giving the solver the most options. Let's see how many ways each step can be solved:

1. 8 (2 by 2 by 2 block on any corner)
2. 3 (extend any side to 2 by 2 by 3)
3. 8 (2 by 2 by 1 can be solved anywhere on either unsolved face)
4. 2 or 4 (next 2 by 2 by 1 block cannot intersect with previous one)
5. 3 (any corner edge pair in a slot)
6. 1? (it's the nice last layer )

This method is even more flexible than petrus (I'm fairly certain, someone correct me if I'm wrong). Steps 3 and 4 remind me of heise's first couple steps, where blocks are made then put into place.


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## qqwref (Dec 27, 2008)

ostracod said:


> 1. 8 (2 by 2 by 2 block on any corner)
> 2. 3 (extend any side to 2 by 2 by 3)
> 3. 8 (2 by 2 by 1 can be solved anywhere on either unsolved face)
> 4. 2 or 4 (next 2 by 2 by 1 block cannot intersect with previous one)
> ...



The way I do it the numbers are actually 8 3 4 3 3 1. For step 3 it is because I want the block to line up so I don't have to do a pseudo-F2L or pseudo-LL, and for step 4 I don't want to have the block cover the 'slot' in the F2L. I guess in theory there are a few other choices but they are a bit unweildy. Remember that it's really important that after step 4 all of the 7 unsolved pieces are touching each other.


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## ostracod (Dec 27, 2008)

You could solve the first 2 layers so that the tripod and last layer are always in place, but I find it to be quicker to take advantage of already formed pairs and make easier blocks. Doing so requires a simple conjugate later on, but it might make recognition for algorithms more difficult. I don't get confused by conjugates for the last step... I guess it's a matter of preference. I am returning home tomorrow.


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## bluephoenix1993 (Dec 27, 2008)

to ostracod

wow, thank you for that tutorial at least i can study your method.. I just want to have another method but as i can see this cannot be used for speedsolving so i guess i'm just gonna practice it to increase the number of methods i can use in solving the rubik's cube.. Well, pretty cool... Thank You...​
to qqwref

Uhm. Can you make a tutorial video or a sample solve of that "tripod F2L method"?.. Well, guess I'm just good at learning when i watch videos... I can't fully understand what you meant... Well.. Thank you...​


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## ostracod (Dec 27, 2008)

Bluephoenix, I know you're probably upset that the supercritical method isn't so fast... I am too! But this method helped me a lot to learn how to intuitively solve. It's especially helpful when learning the tripod method, because most skills can be carried over. If Qqwref won't make a video, then I will do so.


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## qqwref (Dec 27, 2008)

I don't do tutorial videos. Go ahead and do one if you want, ostracod.


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## bluephoenix1993 (Dec 28, 2008)

uhmmm... thank you ostracod... well.. I really liked your method... It was so cool for me because of blockbuilding... Well, somehow, I'm currently learning your method... Uhm... For the tripod F2L video... I guess qqwref won't make one... So I guess can you make one? Please.. Thanks


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## ostracod (Dec 29, 2008)

Alrighty, I made another little site. This one shows how to do the Tripod method in greater detail:

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

Hopefully I'll make a video to show each step.


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## ostracod (Jan 4, 2009)

Here's a video for the Tripod method:

http://www.youtube.com/watch?v=1na2FsB7ThI

Sorry for the long wait!


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## krazedkat (Jan 6, 2009)

Sounds kind of like my combo method (Froux) that I came up with awhile ago....
1. Build 1x2x3 block on LEFT
2. Build 1x2x3 block on RIGHT
3. Finish bottom two edges
4. Solve last layer...


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## fanwuq (Jan 6, 2009)

krazedkat said:


> Sounds kind of like my combo method (Froux) that I came up with awhile ago....
> 1. Build 1x2x3 block on LEFT
> 2. Build 1x2x3 block on RIGHT
> 3. Finish bottom two edges
> 4. Solve last layer...



Bad idea and definitely not original and nothing similar except naming to the above Fridrus method.
It was discussed and shot down a while ago. Pure Roux or Pure Fridrich are more efficient [insert number smaller, but very close to 100]% of the time.


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## krazedkat (Jan 6, 2009)

I know... I don't use that method.. I use Roux... ...


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## qqwref (Jan 6, 2009)

Hi, I've been working on tripod a bit and I thought of a good way to do the LL. It's more efficient than your approach, I think, and it does it in two looks maximum, which is pretty good if you don't want to learn every alg. (By the way I'm not really fond of your "three edges" start, it is a very inefficient way to do a 2x2x3 block. Just do it Petrus style, or even all at once if you can see it.)

First you have to learn the clockwise and counterclockwise "supertwistflip", a pretty short algorithm which twists the three corners in the same direction and flips both edges (of the LL). The ccw one is U' R U R2' F R F2' U F U' and the cw one is the inverse. You can learn fingertricks to make these really fast later. So now when you get to LL you recognize the edges first:
* Case 1: edges solved. You either have a commutator or some kind of corner twist case, so do it intuitively in one step.
* Case 2: edges permuted, but flipped. Use either supertwistflip to go to Case 1.
* Case 3: edges swapped, and maybe flipped. One corner will be correctly permuted, so find it, and if it's twisted apply the correct supertwistflip to fix it. So now you have two unsolved edges (possibly flipped) and two unsolved corners (possibly twisted). This is relatively easy to set up into a PLL, which should be J, R, V, or Y. Here's what you do: set up the corners first (do nothing if they are oriented, otherwise move one of them [without messing up the edges] so it is oriented and still on the LL), and then fix the edges with one move of the M or S slice. Then do your PLL (making sure to swap the pieces' new positions, not the old ones) and undo it all.

Here's an example: set up with D' R U R' U' D R' U R' U' R' U R2 U' y. This is Case 3, and the UBR edge is twisted, so apply the clockwise supertwistflip. Now you have something which looks like a J perm except that all four pieces are twisted. The setup is (corners) L'B' (edges) M', to a J perm. So the entire thing is:
U F' U' F2 R' F' R2 U' R' U
L' B' M' (R' U2 R U R' U2 L U' R U L') M B L.


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## ostracod (Jan 8, 2009)

To all: I am working on a ONE look last layer system which deals with all 108 cases (including mirrors). I will post more info soon.


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## ostracod (Jan 8, 2009)

And yes, qqwref, I probably should get into the habit of doing a petrus style 2 by 2 by 3 block solve. I don't use the 3 edge cross anymore, but I've still been doing it by placing corners first then edges. I think I have an easier time recognizing what to do fast (it takes me 15 seconds to build a block in this way), but I know petrus style would be much faster. New years resolution...


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## qqwref (Jan 8, 2009)

ostracod said:


> I don't use the 3 edge cross anymore, but I've still been doing it by placing corners first then edges.



Corners first? Ewwwwwww... Just try the Petrus way. It'll be hard at first, but if you practice it becomes easier and easier. I just did an average of 5 of 7.81 seconds to build it and I'm not even good at Petrus. I'm sure it's way more efficient than what you're doing anyway. If you want to be a slow intuitive solver you should concentrate on efficiency 


About this "one look last layer system"... I know it's doable, and I think Fridrich's website already has algorithms for all cases (and if she doesn't I *know* Helmstetter's site does), but there's no way I would learn them all. My LL proposal above was for doing it in as few steps as possible without learning any new algorithms.


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## ostracod (Jan 8, 2009)

Yes, I know it's despicable that I do a corners first block; it's a remnant from when I started using the Petrus method as my second method. As a more experienced cuber I should break such habits.

A 1 alg last layer is considerably easier than Fridrich's site would suggest, because with our nice last layer (<-this should get some sort of abbreviation...) it is very easy to apply single move conjugates to form new nice last layers. For instance, the simple 3 corner cycle algorithm can normally cover only 2 cases (clockwise and counterclockwise), but when conjugates are used, this number is increased to 6 cases. I would estimate that one would need to learn maybe 30 algorithms, but that is a rough guess. The down side is you have to remember a lot of cases and which conjugates you should use, but this is a lot easier than remembering a lot of algorithms.

Here is an example code + solution:

"FUR:cycle-cB"

The F means that the left corner's top sticker is in front. The U means the right corner's top sticker is on top. The R means that the corner which is normally on the left side is now on the right. A colon separates the case code from the solution code. The name of the algorithm comes next, in this case "cycle" is the simple 3 corner cycle. Lastly, c stands for conjugate, and B means that the solver will turn the back side to form a new nice last layer. After turning the back side, the solver applies the cycle algorithm to the new LL, then turns the back side... well, back. You can try it for yourself if you want.

I will find out the specific number of algorithms when I find a solution to each case. I have already done 40 cases, all of which can be solved with one of 10 algorithms. I have devised a special case identification code, solution code, and a super fast recognition system, which lets the solver look at only 3 pieces in the nice last layer (the corner + edge + corner in front). This is going to be VERY fast!

PS: I think I am a slow solver for 3 reasons:
1. I have the dexterity of a horseshoe crab. I don't know any finger tricks.
2. Intuitive solving requires a lot of practice to master.
3. A 1 look last layer would do me good. XD It would actually shave off around 8 seconds.


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## ostracod (Jan 8, 2009)

PSS: qqwref, is this you?!

http://www.worldcubeassociation.org/results/p.php?i=2006GOTT01

I don't know how many people have the same name, and if you haven't updated the times in your signature recently... but if it is, I bow down to you! XD


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## krazedkat (Jan 9, 2009)

That IS super critical!


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## ostracod (Jan 9, 2009)

I should really start a new thread, because we are not talking about the Supercritical method now, but the Tripod method. :| The methods are vaguely similar, though, with nice last layers and using "Frifri" to solve corner edge pairs. (Supercritical relies on it too much though!)


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## qqwref (Jan 9, 2009)

ostracod said:


> PS: I think I am a slow solver for 3 reasons:
> 1. I have the dexterity of a horseshoe crab. I don't know any finger tricks.
> 2. Intuitive solving requires a lot of practice to master.
> 3. A 1 look last layer would do me good. XD It would actually shave off around 8 seconds.



1. If you get a good cube and learn a bunch of finger tricks, I think you'd be surprised at how fast you could go. If you just want to practice turnspeed, try to do scrambles as fast as possible. 10 seconds or so is a good time 
2. That's not an obstacle, that's a challenge 
3. 8 seconds? Wow. Have you tried my method for doing the last layer?




ostracod said:


> PSS: qqwref, is this you?!
> 
> http://www.worldcubeassociation.org/results/p.php?i=2006GOTT01



Yeah. The times in my signature are unofficial averages of 12, so they're going to differ from the official averages of 5 by a bit (in either direction ).


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## ostracod (Jan 9, 2009)

A challenge would still require more work nonetheless, but I agree, it's less of an "obstacle" as it is a way of gaining a better understanding for the cube while feeling good about "actually" solving it... I know that the Tripod method doesn't completely use intuition (because of the last step), but it's very close.

I timed myself for doing a 2 corner orienting algorithm (one which is somewhat long, but I am very good at doing), and I got about 7 seconds. Add 1 second for recognition, and that's how much longer it would take to do a 2 look last layer. The first algorithm might require a second longer for recognition, but not much.

Here is how long it takes me (right now) to do each step:

Block: 15-17 seconds
2 by 2 by 1 blocks: 15 seconds
Frifri: 7 seconds
Last layer (2 algs): 15 seconds

That adds up to around 53 seconds.


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## ostracod (Jan 11, 2009)

I went on a little tangent this weekend:

http://www.youtube.com/watch?v=Tm-8ydC2wcc

It is an edges first method. I have no idea how efficient it is. Here are the steps:

1. Solve 5 edges in a pseudo block.
2. Fix "bad" edges (like in the Petrus method).
3. Solve 2 pairs of adjacent edges, one on each unsolved face.
4. Solve the remaining 3 edges, fixing a parity if there is one.
5. Solve the 4 corners on one side, using 1 of 2 algorithms for each.
6. Solve the last 4 corners, using 1 algorithm to orient and 1 to permute.


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