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Random Cubing Discussion

Smiles

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Parity means that the states with an odd number of swaps are unreachable, eg if you just swapped 2 edges, then that is one swap and the case is unreachable (without disassembly)

Sorry i deleted my comment because I found out what parity was LOL.
I kinda meant parity in context to Roux, which I think isn't the same thing. I know what parity is in general though, like on 4x4, 6x6, void cube, etc.
According to google it's the E2 M E2 M' case.

The 2 is parity. The 6 is for circular permutations. Basic explanation: UF->UR->UL is the same as UR->UL->UF. So it is really (n-1)! or n!/n

are you sure the circular permutation argument can eliminate cases?
because here's what i'm thinking:
6! means first edge can go anywhere (6 possibilities), next edge can go in anywhere of the 5 remaining spots, etc.
now that has nothing to do with swaps (it's not counting swaps), just the position of each edge. UF -> UR -> UL is one of the 6! possible permutations. UR->UL->UF is the same permutation, it never got counted multiple times anyway so i don't see why you can divide to get rid of it.

anyway i might be wrong, can someone care to explain this?
my math is just L6E cases = 6!/2, or LnE = n!/2
and then there's the solved case, so number of algs is just subtract 1.

intuitively for like L3E there would be 1 solved case + 2 3-cycles, which makes 3 cases or 2 algs.
3!/2 = 3 cases, and subtract 1 is 2 algs.
and then obviously for L2E there would be 1 solved case and 0 algs.
2!/2 = 1 case, and subtract 1 is 0 algs.

shouldn't the same thing work for L6E?

edit: i meant permutation only, like PL6E rather than L6E
 
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yoinneroid

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are you sure the circular permutation argument can eliminate cases?
because here's what i'm thinking:
6! means first edge can go anywhere (6 possibilities), next edge can go in anywhere of the 5 remaining spots, etc.
now that has nothing to do with swaps (it's not counting swaps), just the position of each edge. UF -> UR -> UL is one of the 6! possible permutations. UR->UL->UF is the same permutation, it never got counted multiple times anyway so i don't see why you can divide to get rid of it.

anyway i might be wrong, can someone care to explain this?
my math is just L6E cases = 6!/2, or LnE = n!/2
and then there's the solved case, so number of algs is just subtract 1.

intuitively for like L3E there would be 1 solved case + 2 3-cycles, which makes 3 cases or 2 algs.
3!/2 = 3 cases, and subtract 1 is 2 algs.
and then obviously for L2E there would be 1 solved case and 0 algs.
2!/2 = 1 case, and subtract 1 is 0 algs.

shouldn't the same thing work for L6E?

Nope, you need to multiply further by 2^5 for orientation to calculate L6E
and also there are cases UF-UL-UB and UF-UB-UR which will be counted differently by the 6!, but it can be solved by the same algs for PL6E with some AUF, maybe the only way to remove these is by manual counting? I actually did that for my 1LLL counting :/
 

Smiles

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Nope, you need to multiply further by 2^5 for orientation to calculate L6E
and also there are cases UF-UL-UB and UF-UB-UR which will be counted differently by the 6!, but it can be solved by the same algs for PL6E with some AUF, maybe the only way to remove these is by manual counting? I actually did that for my 1LLL counting :/

oh we're doing EO? i thought CLLEO came before this. was i mistaken? o_o

EDIT: my mistake, i forgot to say PL6E instead of L6E

and your 3-cycle example makes sense. anyway i forgot AUFs. TheNextFeliks's example was describing 2 cases for the same edge permutation. idk maybe i just didn't get it :p
 
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goodatthis

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Has anyone ever attempted GigaminxBLD?

In terms of number of pieces, it should be comparable to 8BLD or 9BLD. Not sure how the execution would go.
Yep. I asked this a while ago, I believe mike hughey has attempted it. I think he said he was off by 20%.

edit:

I've tried Gigaminx BLD three times so far (quite a while ago); attempts were 6 hours the first time, 4 1/2 hours the second and third time. All were around 20% incorrect. I still have my gigaminx on the kitchen counter staring at me, making me feel guilty every time I go by it and don't pick it up to try again.


Oh, and Chris Hardwick and I have both done 4/4 4x4x4 multis in negative time (2010 and 2011). And while we're talking about that, last year I had a -17 minute 7x7x7 BLD solve.


I had a 3/4 5x5x5 multi a lot of years ago. And a 1/2 7x7x7 multi. The latter one is on my youtube channel.
 
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yoinneroid

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oh we're doing EO? i thought CLLEO came before this. was i mistaken? o_o

EDIT: my mistake, i forgot to say PL6E instead of L6E

and your 3-cycle example makes sense. anyway i forgot AUFs. TheNextFeliks's example was describing 2 cases for the same edge permutation. idk maybe i just didn't get it :p

Actually, I would like to keep it at 359/360 cases, treating cases that need AUF as different algs, the existence of another 'free' layer (M layer) complicates things a bit too much for me :p
 

cmhardw

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I've always wondered if someone stopped cubing for a long time, would they remember their algorithms after coming back to it again? I recently discovered something about myself: I forgot the 4x4x4 wing edge parity algorithm that I first learned.

I was able to recall it by experimenting on the cube some, using a combination of knowing what exactly a parity algorithm is trying to achieve (scramble and solve centers using an odd number of inner slice quarter turns), as well as using muscle memory. I was able to, without looking it up, remember my old parity algorithm!

Still, I think it's weird that I can forget an alg I've executed hundreds of, perhaps even a couple thousand, times over the years. So I think the answer is: after extended breaks you do forget algorithms. However, they can be brought back if you can find one of your old associations to that alg. Neato? :)
 
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TDM

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I've always wondered if someone stopped cubing for a long time, would they remember their algorithms after coming back to it again? I recently discovered something about myself: I forgot the 4x4x4 wing edge parity algorithm that I first learned.

I was able to recall it by experimenting on the cube some, using a combination of knowing what exactly a parity algorithm is trying to achieve (scramble and solve centers using an odd number of inner slice quarter turns), as well as using muscle memory. I was able to, without looking it up, remember my old parity algorithm!

Still, I think it's weird that I can forget an alg I've executed hundreds of, perhaps even a couple thousand, times over the years. So I think the answer is: after extended breaks you do forget algorithms. However, they can be brought back if you can find one of your old associations to that alg. Neato? :)
I've found I have forgotten the first ever OLL parity alg I learned (the horrible pure edge flip one), but tbh it isn't that important anyway. I can do all COLLs for the U case, so I just do my normal OLL parity (actually double parity) alg followed by COLL, which is probably faster than doing the pure flip alg when I get pure flip for OLL just because the alg is so bad, which could also lead to another bad PLL.
 

tseitsei

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I've always wondered if someone stopped cubing for a long time, would they remember their algorithms after coming back to it again? I recently discovered something about myself: I forgot the 4x4x4 wing edge parity algorithm that I first learned.

I was able to recall it by experimenting on the cube some, using a combination of knowing what exactly a parity algorithm is trying to achieve (scramble and solve centers using an odd number of inner slice quarter turns), as well as using muscle memory. I was able to, without looking it up, remember my old parity algorithm!

Still, I think it's weird that I can forget an alg I've executed hundreds of, perhaps even a couple thousand, times over the years. So I think the answer is: after extended breaks you do forget algorithms. However, they can be brought back if you can find one of your old associations to that alg. Neato? :)

Seems about right.

Nowadays I solve 3x3 so little that I sometimes forget some of my OLLs, but I can usually "get them back" by messing around with the case for a while and just remembering that alg again...
 

Renslay

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I've always wondered if someone stopped cubing for a long time, would they remember their algorithms after coming back to it again? I recently discovered something about myself: I forgot the 4x4x4 wing edge parity algorithm that I first learned.

I was able to recall it by experimenting on the cube some, using a combination of knowing what exactly a parity algorithm is trying to achieve (scramble and solve centers using an odd number of inner slice quarter turns), as well as using muscle memory. I was able to, without looking it up, remember my old parity algorithm!

Still, I think it's weird that I can forget an alg I've executed hundreds of, perhaps even a couple thousand, times over the years. So I think the answer is: after extended breaks you do forget algorithms. However, they can be brought back if you can find one of your old associations to that alg. Neato? :)

I had to re-learn some 4x4 and square-1 algs after I didn't solve them for months - despite I used those algorithms many hundred times.
Sometimes I was able to reconstruct the algs, and sometimes I had to look after them, which caused a "Heureka! Now I remember!" moment. :)
 

Mike Hughey

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Still, I think it's weird that I can forget an alg I've executed hundreds of, perhaps even a couple thousand, times over the years. So I think the answer is: after extended breaks you do forget algorithms. However, they can be brought back if you can find one of your old associations to that alg. Neato? :)

I've forgotten a number of algorithms I've executed thousands of times WITHOUT having taken an extended break. Unless you call a week an extended break. It's very frustrating - I think that's why I have so much trouble getting very good at 3x3x3 speedsolving.

Fortunately, like you say, it is usually possible to eventually dredge it up, although sometimes it is quite hard.

I've forgotten my 2-bar-in-front PBL for 2x2x2 5 or 6 times - one time it happened to me at Nationals, after I had done several hundred 2x2x2 solves in practice the week before the competition. I remember I had to ask Shaden to show it to me right before I went up to do 2x2x2. I'm sure I must have done that alg thousands of times.

I often wonder if this is age-related.
 

CyanSandwich

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I've always wondered if someone stopped cubing for a long time, would they remember their algorithms after coming back to it again? I recently discovered something about myself: I forgot the 4x4x4 wing edge parity algorithm that I first learned.

I was able to recall it by experimenting on the cube some, using a combination of knowing what exactly a parity algorithm is trying to achieve (scramble and solve centers using an odd number of inner slice quarter turns), as well as using muscle memory. I was able to, without looking it up, remember my old parity algorithm!

Still, I think it's weird that I can forget an alg I've executed hundreds of, perhaps even a couple thousand, times over the years. So I think the answer is: after extended breaks you do forget algorithms. However, they can be brought back if you can find one of your old associations to that alg. Neato? :)
I still do square-1 every week for the weekly comp, but I sometimes forget algs. Although I generally remember them without looking them up. I also learned some corner comms as algs so I could use TuRBo corners for 3BLD. But I didn't use them enough and forgot most of them.

I've forgotten all the gear cube algs. I suppose I've forgotten all the Dan Brown 3x3 algs that weren't transferred to CFOP.
 

IRNjuggle28

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When you got your first sub 10 single, what was your average of 100 PB? I have a 14.78 AO100, and I still don't have a sub 10. I've come very close, and failed a lot of fast solves.
 

guysensei1

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When you got your first sub 10 single, what was your average of 100 PB? I have a 14.78 AO100, and I still don't have a sub 10. I've come very close, and failed a lot of fast solves.

First sub-10 single: 9.91 LL skip when I was averaging ~17-18

First fullstep sub-10: 9.xx single when I was averaging low 14. I don't do ao100s very often at all, but I would say it would be around low 14 as well.
 

XTowncuber

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When you got your first sub 10 single, what was your average of 100 PB? I have a 14.78 AO100, and I still don't have a sub 10. I've come very close, and failed a lot of fast solves.

I bet I had a sub 10 ao12 before I ever actually did an ao100.

But yeah, I probably averaged 14-15 when I got my first one. You'll get it soon, I'm sure.
 

guysensei1

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I just realised that this particular N perm for OH:
R U R' U R U2 R' U' R U2 L' U R' U' L U' R U' R'
is actually just sune + some ZBLL.

That ZBLL section is actually pretty nice for diagonal swap antisune COLL.
 

ryanj92

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When you got your first sub 10 single, what was your average of 100 PB? I have a 14.78 AO100, and I still don't have a sub 10. I've come very close, and failed a lot of fast solves.
I wanna say it was high 16, maybe very low 17. 8 move LL though :p
(That was about 9 months ago, and I've only had one more since, which was 'full step' (2 move cross, obvious pairs, LL was sune into T perm or something similar). I also fail a lot of easy solves...)
 

Bindedsa

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I just realised that this particular N perm for OH:
R U R' U R U2 R' U' R U2 L' U R' U' L U' R U' R'
is actually just sune + some ZBLL.

That ZBLL section is actually pretty nice for diagonal swap antisune COLL.
For COLL, I prefer: R' U' R U' L U' R' U L' U2 R. It's just the inverse of the FB mirror to the case you referenced. I never used it though, just did sune until I finished the set.
 
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