Merging step 5 into step 3 in Petrus

Rubik's Noob

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Hi everyone!

I've found that if you take a solved cube, scramble just two layers, and then start the Petrus method at step 4, you never have to do step 5. That means the corner positions are fixed after step 3. (This is probably widely known, right?)

So I've come up with a method to check the corner positions before fixing the last two bad edges in step 3, and to fix them along with the bad edges, using very few additional moves (usually around 1 or 2; maximum is 5, I think). It's not easy, and I can't do this very fast yet, but I've just started solving Rubik's cubes anyway, so others can probably do it much more quickly. The other aspect is that if you never have to do step 5, you can learn how to do steps 6+7 in one go, and forget about 5+6.

Just wondering: Is this old news? If not, would anyone be interested in a tutorial? And if so, are there any freely available resources (e.g., Java cubes) for making one?
 
Steps 6+7 is 2GLL?

Anyhow, this hasn't seriously been tried very much. Recognition is probably not worth it, because most people don't do Petrus LL as his steps 5+6+7.

But I'm very interested in any good work you have on this.

Try this for an applet.
 
This method I think was come up by FMC solvers some years ago. Sebastian D. (can't be bothered to spell his surname) had all the necessary algorithms.
 
After a few seconds of searching using the name provided by Timothy I was able to find this page where Sebastian describes his method for bringing the cube into a state that can be solved with turns of only two faces.
http://www.geocities.com/portoseb/cube/method.html
I still find it hard (or impossible) to tell which two corners need to be switched but at least now I have some cases and algs..
 
Too late :-)

Thanks for the replies. Especially thanks to Lucas Garron for the link. I guess I didn't find anything new then, but since my tutorial is already half-finished now, I'll post a link here anyway:

http://www.stud.uni-karlsruhe.de/~uxauu/Rubik/

In the Java cubes, when you loop over the end of the sequence, the sequence starts again, but the cube is not reset. Just so you don't get confused...

The part that I have not written yet is how to check the permutation without making any moves. Unfortunately, that is also the part that I cannot do quickly. The rest is no problem, even though my description may be a bit lengthy.
 
I was the first to use this method, probably like 4-5 years ago. So it's not very new :) I came up with it while doing the fewest moves challenges, I still use it to find decent solutions. But I couldn't ever convince people to use this method as well. Sebastian was the only other person to ever solve this way but he didn't learn from my pages, he discovered it on his own (I used to have web pages describing the method, I had the cases for 4 and 6 bad edges also. But I removed the pages after most people were finding them by searching for something like "very easiest method" and weren't clicking through to the pages with the algorithms)

Anyway you can see a bunch of my solutions from the FMC archives using this method at

http://www.cubestation.co.uk

or at

http://www.ryanheise.com/competitions/linearfmc/rankings.cgi
(select the 1st tournament from the menu. I used this method almost exclusively here)

Anyway, this method will never be used for speedsolving since recognition takes too long.
 
Very nice. I enjoyed reading it.

Thank you!

I figured quick recognition would be the most difficult part. How have you done it thus far?

I have streamlined my method of recognizing the permutation, and added it to the tutorial. Don't be scared by the length of the explanation. At the end, when I describe how this applies to the example, you may find that with some practice, it does not have to take that long.
 
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