One more suggestion: No BB2 state can be transformed into Bars/One Bar directley. ...Only the BB1 can. that means this connection on the extended maze is wrong. Or I am wrong... It was just important to me, because, i was going to add this connection to my simplified Maze. I would be appreciative for a disconfirmation/confirmation just to be sure.
Awesome! I'm comfortable using DR in my FMC solves, but I'm still learning to use HTR. This video and your document will certainly help me a lot :) Thanks!
So I've been playing around with this method for a couple of days, and I'm currently getting ~30 move solutions. I can imagine this will quickly get better, as I'm quite slow right now and one hour is not enough time for me to explore a lot of different hyper-parity continuations after DR. But this is promising!
I have been trying to learn how to use htr, but there is very little information out there about how i am supposed to learn htr, such as where to start. i tried using this as a guide and trying to go from step step, but that didn't work.there are a few things i didn't understand. How to count quarter turns needed and how to memorize how i am supposed go along the different routes
It's a good case if you can split up the two bad F/B stickers (1 on U and 1 on D) while still forming a bar/slash pattern with the U/D stickers. This sets up a transition to bars + bars, which is impossible in the bad case.
Can you explain how to recognize "same" and "exclusive" UD and FB bars? I think I understand what you mean by "different", however, the way I recognize it is the white stickers match with all 4 bad or good FB pieces (which to me seems counterintuitive because I'd call that case "same").
In the "exclusive" case, if you try to form bars with the the bad FB stickers, you will find that the bad UD stickers are split 3/1 between the top and bottom faces, and vice versa. So it's not possible to form bars with both the bad UD and the bad FB stickers simultaneously. The good case is named "same" because you can form both sets of bars simultaneously, with the bad UD and the bad FB stickers on the same cubelets
Hi Rodney. Fantastic video! I have watched this a few times now and I think the concepts you have covered are fascinating. This is one of coolest things I have learnt in cubing in a while. At the moment, I am learning to differentiate between: ST1, ST2 BB1, BB2, BB3 BS1, BS2 I'm just wondering what ST1 ST2 and BS1 BS2 stand for? I was wondering why not use something like SOF and BSOB (although these acronyms are longer)
Those names are supposed to identify the points in the maze. ST = Solved/Top. BB = Bars/Bars, BS = Bar/Slash. Not very clear, I admit. Maybe I should have just gone with A/B/C!
... i might just have dismantled it myself. I think, when the two perpendicular Bars are in one layer, U doesn't affect anything.... So is the inverse true? - Any BB2 can be transformed into 2B/2B directley? It seemed impossible to me in the situation i just encountered.
I have a question/hypothesis: is it true, that the BS2 (Bar/Slash - One Bar, Perpendicular) has actually two distinct sub types: One can go to 2B/1B and the other not? (no mather how many preparing halfturns are done) So in the BS2 state a U-quater-Turn can either switch between this two BS2-states or switch to the BS1 state? More precise: if the perpendicular Bars are beeing on the same Layer, it stays within the BB2.
If the perpendicular bars are on the same layer, then a U move stays at BS2, but there's only one BS2. If you're having trouble finding the half-turn setups that take you to somewhere that the maze says you should be able to go to, then R2 U2 F2 will often work.
Thank you for creating the hyper pairity maze! I love it but sometimes i get lost in it. so this morning i run into a situation, that leads to a question: in the constellation UD=two bars combined with FB=one face, i think, i found out, that there is also a good and a bad situation, similar to the three that you mention in your document. The bad case can only go to UD=bar slash and FB=two bars, whereas another can go to UD=solved and FB=One face as well as to Cornersolution solved directly. is this true or is there an error in my thougts? am really comitted to learn it!
This morning, I found the origin of this issue problem myself. After contemplating for three days that there must be another special case with both a good and a bad option, I discovered that the supposedly bad option I was dealing with was physically not possible! :-) This was due to the fact that I had only mentally considered those cases. I had set it up just in my mind without considering whether the situation could actually occur in reality. I am relieved that the mystery is solved."
I am *really* willing to learn this and see how it goes. The facts at the end are kinda shocking. Will see what I can do. Thanks for this!!!
Awesome. Let me know how things go!
This is mind blowing
One more suggestion:
No BB2 state can be transformed into Bars/One Bar directley.
...Only the BB1 can. that means this connection on the extended maze is wrong. Or I am wrong...
It was just important to me, because, i was going to add this connection to my simplified Maze. I would be appreciative for a disconfirmation/confirmation just to be sure.
Awesome! I'm comfortable using DR in my FMC solves, but I'm still learning to use HTR. This video and your document will certainly help me a lot :) Thanks!
So I've been playing around with this method for a couple of days, and I'm currently getting ~30 move solutions. I can imagine this will quickly get better, as I'm quite slow right now and one hour is not enough time for me to explore a lot of different hyper-parity continuations after DR. But this is promising!
Where did you learn DR
@@alexting827 Mainly using drive.google.com/drive/folders/1mppifILqu9Bu2phr8zhXGcXasBsSkv_S?usp=sharing
I have been trying to learn how to use htr, but there is very little information out there about how i am supposed to learn htr, such as where to start. i tried using this as a guide and trying to go from step step, but that didn't work.there are a few things i didn't understand. How to count quarter turns needed and how to memorize how i am supposed go along the different routes
hey, thanks for the video! how can i easily identify if it is a good or a bad case on the One Bar Bar/Slash
It's a good case if you can split up the two bad F/B stickers (1 on U and 1 on D) while still forming a bar/slash pattern with the U/D stickers. This sets up a transition to bars + bars, which is impossible in the bad case.
Can you explain how to recognize "same" and "exclusive" UD and FB bars? I think I understand what you mean by "different", however, the way I recognize it is the white stickers match with all 4 bad or good FB pieces (which to me seems counterintuitive because I'd call that case "same").
In the "exclusive" case, if you try to form bars with the the bad FB stickers, you will find that the bad UD stickers are split 3/1 between the top and bottom faces, and vice versa. So it's not possible to form bars with both the bad UD and the bad FB stickers simultaneously. The good case is named "same" because you can form both sets of bars simultaneously, with the bad UD and the bad FB stickers on the same cubelets
Hi Rodney. Fantastic video! I have watched this a few times now and I think the concepts you have covered are fascinating. This is one of coolest things I have learnt in cubing in a while.
At the moment, I am learning to differentiate between:
ST1, ST2
BB1, BB2, BB3
BS1, BS2
I'm just wondering what ST1 ST2 and BS1 BS2 stand for? I was wondering why not use something like SOF and BSOB (although these acronyms are longer)
Those names are supposed to identify the points in the maze. ST = Solved/Top. BB = Bars/Bars, BS = Bar/Slash. Not very clear, I admit. Maybe I should have just gone with A/B/C!
... i might just have dismantled it myself. I think, when the two perpendicular Bars are in one layer, U doesn't affect anything....
So is the inverse true? - Any BB2 can be transformed into 2B/2B directley?
It seemed impossible to me in the situation i just encountered.
I have a question/hypothesis: is it true, that the BS2 (Bar/Slash - One Bar, Perpendicular) has actually two distinct sub types: One can go to 2B/1B and the other not? (no mather how many preparing halfturns are done)
So in the BS2 state a U-quater-Turn can either switch between this two BS2-states or switch to the BS1 state? More precise: if the perpendicular Bars are beeing on the same Layer, it stays within the BB2.
If the perpendicular bars are on the same layer, then a U move stays at BS2, but there's only one BS2. If you're having trouble finding the half-turn setups that take you to somewhere that the maze says you should be able to go to, then R2 U2 F2 will often work.
Thank you for creating the hyper pairity maze! I love it but sometimes i get lost in it.
so this morning i run into a situation, that leads to a question:
in the constellation UD=two bars combined with FB=one face, i think, i found out, that there is also a good and a bad situation, similar to the three that you mention in your document. The bad case can only go to UD=bar slash and FB=two bars, whereas another can go to UD=solved and FB=One face as well as to Cornersolution solved directly.
is this true or is there an error in my thougts?
am really comitted to learn it!
This morning, I found the origin of this issue problem myself. After contemplating for three days that there must be another special case with both a good and a bad option, I discovered that the supposedly bad option I was dealing with was physically not possible! :-) This was due to the fact that I had only mentally considered those cases. I had set it up just in my mind without considering whether the situation could actually occur in reality. I am relieved that the mystery is solved."
Well that a relief that there are no holes in this method
Also I want to know whether DR corner solution or hyper parity was better (fewest move), or they equal and I can pick whatever I want
Hey! This method is only to solve corners after DR right? Then you would need to apply insertions for edges?
That's right. I have another guide coming out that will cover edges.
Imagine you are old and a little feeble minded. Is there an easier place to start learning this fascinating but cryptic material?
Check out docs.google.com/document/d/1_NMlyPh2Ifd1E0ivos-15Sp-by-KPxoOsj9JD4pJuHI/edit#heading=h.3km7zdqm21jw