# Helicopter cube solving (and patterns)



## mrCage (Apr 29, 2010)

Hey all!

I just wonder what method you guys use for helicopter cube solving. I just got mine today, and i do corners first solving quite easily - though not fast yet.

I have also been pondering about various patterns possible on this puzzle. One obvious one is the "2 spot pattern". Interchange all the face centers of 2 opposite face centers. I can do this in 24 turns, not all that efficient. The cube in cube pattern seems impossible. Can anyone prove this??

Per

PS! My helicopter cube (from Mefferts) does not seem to have shape shifting turns. Is this correct??


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## qqwref (Apr 29, 2010)

There are shape shifting (jumbling) turns but they aren't obvious. Slowly turn one axis until an adjacent one will turn and you will see how it works. You can't really move corners out of position (except temporarily) but you can move centers into different orbits. So any pattern with solvable corners is possible.

My method (when it's been scrambled without jumbling) is as follows:
- First layer centers.
- Finish first layer by pairing the corners up with two centers each before inserting.
- Last layer corners (I solve one by one, then do last two at once).
- Last layer centers, do the 3-cycles first and then any 2-cycles in pairs.
I use three main algs: J-perm (UL UF UR UF UL), 2-2 cycle ((UF UR)3), 3-cycle ((UF FR UR FR)2) and inverse. And see, it works pretty well:
36.48, (1:01.38), (34.00), 51.96, 42.26, 41.95, 46.47, 57.55, 54.00, 45.83, 49.07, 47.68 => 47.32


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## mrCage (Apr 30, 2010)

qqwref said:


> There are shape shifting (jumbling) turns but they aren't obvious. Slowly turn one axis until an adjacent one will turn and you will see how it works. You can't really move corners out of position (except temporarily) but you can move centers into different orbits. So any pattern with solvable corners is possible.
> 
> My method (when it's been scrambled without jumbling) is as follows:
> - First layer centers.
> ...


 
Hi. I have to sit down and contemplate over some diagrams to fully understand your method. Solving without shape shifting turns is always possible anyhow-evn if scrambled with them.

And yes i realised that cube in cube pattern is indeed possible. I managed to twist 2 wings (with 2 easy corner 3-cycles). And the centers are indeed possible to set up also.

Per


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## TomZ (Apr 30, 2010)

mrCage said:


> Solving without shape shifting turns is always possible anyhow-evn if scrambled with them.


That's not true. There are many positions that you cannot achieve without jumbling.


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## qqwref (Apr 30, 2010)

mrCage said:


> And yes i realised that cube in cube pattern is indeed possible. I managed to twist 2 wings (with 2 easy corner 3-cycles). And the centers are indeed possible to set up also.


It's not obvious that the centers can be setup in this way - at least, it's impossible without jumbling. Using non-jumbling moves there are four orbits (Urf Rbu Bdr Dlb Lfd Ful, and the same with y/y'/y2 rotation), each with one centers of each color, which the centers can't move out of, although with jumbling moves you can move the centers around freely.


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## mrCage (May 1, 2010)

TomZ said:


> mrCage said:
> 
> 
> > Solving without shape shifting turns is always possible anyhow-evn if scrambled with them.
> ...


 
I think we would have to distinguish between normal helicopter cube and a picture version where the exact location of centers had to be fixed.

Center swap is possible when combined with another invisible swap...

Per


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## mrCage (May 1, 2010)

qqwref said:


> There are shape shifting (jumbling) turns but they aren't obvious. Slowly turn one axis until an adjacent one will turn and you will see how it works. You can't really move corners out of position (except temporarily) but you can move centers into different orbits. So any pattern with solvable corners is possible.
> 
> My method (when it's been scrambled without jumbling) is as follows:
> - First layer centers.
> ...


 
I was able to create an adjacent 2 spot pattern by using combination of your 2 given center algs: UF FR UR FR UF UR FR UR, a dual swap. Use the symmetrical version to create the pattern.

I can also twist 3 non-adjacent wings (corners) in 16 turns.

Per


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## bobso2 (May 1, 2010)

where is a good place to buy this puzzle?


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## mrCage (May 1, 2010)

bobso2 said:


> where is a good place to buy this puzzle?


 
I got mine from Mefferts. Another good place is the Twisty store. Not sure which is faster:confused:

Per


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## cubesolver77 (May 1, 2010)

I would recommend the twisty store even though their is a wait that is it was originally made but i jumped too soon and ordered from mefferts without ordering it


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## jaap (May 1, 2010)

mrCage said:


> TomZ said:
> 
> 
> > mrCage said:
> ...



No. It is impossible to swap two centre pieces of the same colour, without shapeshifting/jumbling.
It is also impossible for example to make the top-right centre piece in the front face and the front-left centre piece in the top face the same colour.
Choose a single centre piece, and see where you can place that piece using only non-jumbling moves. It cannot be placed anywhere - it is very restricted.


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## mrCage (May 1, 2010)

I can swap 2 corners like so:

1, swap some edges:
UR UF FR (UR UF)*3 FR UF UR

2, swap the same edges with single turn
FR

Per


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## Portponky (May 1, 2010)

I made a guide for solving and unjumbling the helicopter cube. I think it's pretty similar to a lot of what is discussed here, so it may not be too helpful.

It is very easy to jumble in to a position that can't be solve without jumbling again. Just start from solved, do a single jumble move and try to solve it without re-jumbling.


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## mrCage (May 1, 2010)

jaap said:


> mrCage said:
> 
> 
> > TomZ said:
> ...


 
Yes i see it now. But i still dont see the jumbling move. I can half see it, but to complete it i seem to have to undo it. Hmmm. I guess i'll just live without them

Per


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## mrCage (May 2, 2010)

Hi!

Corner 3-cycle:
(FR FD FR FU)*2

Now use this to create a 2-twist on corners: (UFR-,UFL+)
1: (FR FD FR FU)*2 

followed by a symmetric version
2: (FD FR FD FL)*2

Per

PS Double sune 2-twist on rubiks 333 springs to mind!!


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## mrCage (May 3, 2010)

Corner 3-twist:

((FR FD FR FU)*2 FL)*2 (18 turns) Is this optimal??

Per


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## Portponky (May 3, 2010)

That's a cool move. I don't know if it is optimal.

I think you can solve the last layer corners in 19 twists maximum, but it will mix up the centers on the top half.


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## mrCage (May 4, 2010)

Portponky said:


> That's a cool move. I don't know if it is optimal.
> 
> I think you can solve the last layer corners in 19 twists maximum, but it will mix up the centers on the top half.


 
Any reasoning for the 19 maximum. What are the steps ...

Per


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## coinman (May 4, 2010)

mrCage said:


> Corner 3-twist:
> 
> ((FR FD FR FU)*2 FL)*2 (18 turns) Is this optimal??
> 
> Per



I found that (UR UB UL UF) *3 makes a 3 corner twist without mowing any centers.
Also (UR UL UB UF)*3 moves four corners without mowing any centers.

I got mine yesterday and solved it in about there hours without help from the internet but now i will only end up in strange positions on the last centers that i can't solve with the algs i found this far. I made a fev jumbling moves, maybe it's in unsolvable position.


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## Portponky (May 4, 2010)

coinman said:


> I got mine yesterday and solved it in about there hours without help from the internet but now i will only end up in strange positions on the last centers that i can't solve with the algs i found this far. I made a fev jumbling moves, maybe it's in unsolvable position.



It is very likely that it will require unjumbling to be solvable. The two obvious indicators are:

A center piece is in the same orbit as another center of the same colour. That is, two same-coloured center pieces can both get in to the same position.
You can solve almost all of it but two centers are always swapped.

I posted an unjumbling guide (see post #13), which is a bit rough at the moment but should help you if you are stuck.



mrCage said:


> Any reasoning for the 19 maximum. What are the steps ...



To orient them takes a maximum of two moves. Twist two edges (e.g. UF UR) and the orientation changes in a sune-like way. The permutation changes too but ignore that. Two sunes can orient any corner orientation so 2*2 = 4 moves can orient all four corners.

To permute them I thought the worst case was 15 but it's actually 12 so the corners can be solved in 16 max.

You can split it up in to 8 cases. I'll list the algorithms but because all of them are on the U face *I've written them without the preceeding 'U's*. So from here, FLR means UF, UL, UR.

1. Solved

2. R-style perm, two corners solved, two needing swapped along an edge.
LRBLR

3. E-style perm, two pairs of corners needing flipped along edges.
LRFBLR

Note: The above two moves are enough to solve the corners if you apply them enough times. This is more than enough for speed solving.

For the remaining cases I have constructed example algorithms using a method I'll explain.

4. H-style, where there is an hourglass like cycle (BL -> BR -> FL -> FR -> BL).
LFBR FBL RBFL

5. X-style, where two pairs of corners are diagonally swapped.
LFBR LFBL RBFL

6. O-style, where the corners are all correct relative to each other but they are a quarter turn out of a alignment with the correct position.
LFBR BLF RBFL

7. Y-style where two diagonal corners are swapped but the other two are solved.
LFBR BLB RBFL

8. A-style, where you have 1 solved and a 3-cycle
LFBR LB RBFL

Counting mirror images and reflections that should solve all corner cases (unless I've stupidly missed one) and the worst is the X-perm which takes 12.

The majority of those algorithms are wrapped in LFBR and RBFL. That's a handy setup move which puts it in a situation where you can make a lot of swaps without breaking the orientation. F and B swap the corners they usually would but without breaking the orientation, and L swaps the corners on the R edge without breaking orientation. R will break the orientation so you're restricted there.

If you can see how to permute the corners using F, B and R twists then just do LFBR, then do the twists you saw but using L instead of R, and then do RBFL.

Of course all of this will mess up the centers on the top half of the cube so it may not mesh well with your method.


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## coinman (May 4, 2010)

Just got a 5:51.xx solve on my third attempt (the two first was hoers long). I bought two of them so i still have one that is solvable


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## qqwref (May 4, 2010)

I wonder what the God's Algorithm count would look like for LL corners, just using UL/UB/UF/UR turns.

Well, actually, nevermind that, I did the calculations. Here are the number of positions at each depth:
0 moves: 1 (that's the solved case)
1 move: 4
2 moves: 10
3 moves: 20
4 moves: 34
5 moves: 52
6 moves: 74
7 moves: 100
8 moves: 130
9 moves: 128
10 moves: 74
11 moves: 20
12 moves: 1 (that's the U2 case)
The average case takes 7.39 moves to solve.


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## Portponky (May 5, 2010)

Here is a 9 move Y-perm for the last layer: UR, UB, UR, UL, UF, UL, UR, UB, UR.


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## mrCage (May 5, 2010)

qqwref said:


> I wonder what the God's Algorithm count would look like for LL corners, just using UL/UB/UF/UR turns.
> 
> Well, actually, nevermind that, I did the calculations. Here are the number of positions at each depth:
> 0 moves: 1 (that's the solved case)
> ...


 
Details??

Per


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## qqwref (May 5, 2010)

What do you mean? What details do you want? I'm not gonna just go posting an optimal solution to each position


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## mrCage (May 5, 2010)

qqwref said:


> What do you mean? What details do you want? I'm not gonna just go posting an optimal solution to each position


 
I mean how did you find it? GAP??

Per

I came to think of an interesting subgroup: [UF,UR,FR] - a 3 generator subgroup. What would gods algorithm be for this group? Rather "Gods distribution" 

Per


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## qqwref (May 5, 2010)

No, I don't have GAP, I used ksolve (+ some javascript stuff to generate cases).

The <UR,UF,FR> subgroup sounds interesting, only about 2600 cases so it's still doable with my primitive method  I'll take a look.

EDIT: Tried it, here's what comes up:
Depth 0: 1
Depth 1: 3
Depth 2: 6
Depth 3: 12
Depth 4: 24
Depth 5: 48 [it stops doubling here, so if you do up to 5 moves there is exactly 1 optimal solution and it's the inverse]
Depth 6: 93
Depth 7: 180
Depth 8: 315
Depth 9: 489
Depth 10: 604
Depth 11: 522
Depth 12: 250
Depth 13: 42
Depth 14: 3 [RF UF RF UR UF UR UF RF UF RF UR UF UR UF, or a corner 2-2 cycle, and its rotations]
So that's 2592 total cases, with an average of 9.45 moves to solve.


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## coinman (May 5, 2010)

qqwref said:


> My method (when it's been scrambled without jumbling) is as follows:
> - First layer centers.
> - Finish first layer by pairing the corners up with two centers each before inserting.
> - Last layer corners (I solve one by one, then do last two at once).
> ...



How do you do this step? I can't find a way to pair the corners with centers easily and if manage i can't keep them paired up wile moving it to insert it in the right slot.


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## Portponky (May 5, 2010)

coinman said:


> qqwref said:
> 
> 
> > - Finish first layer by pairing the corners up with two centers each before inserting.
> ...



What I do is first I get the corner underneath its final location with the top layer colour facing downwards, so it's only one twist away from being solved. You want to get one center on either side of this corner. Find one of the centers (there are only four valid locations) and get it one twist away from aligning with the corner. Twist the corner away, twist the center in, and then twist the corner back. Repeat this for the other center, and then you can twist the corner up to the top layer along with the centers. Repeat again for the remaining corners and you've done that step.


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## mrCage (May 5, 2010)

qqwref said:


> No, I don't have GAP, I used ksolve (+ some javascript stuff to generate cases).
> 
> The <UR,UF,FR> subgroup sounds interesting, only about 2600 cases so it's still doable with my primitive method  I'll take a look.
> 
> ...


 
How do you make ksolve run so fast?? Hmm...

Per


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## qqwref (May 5, 2010)

Hm, what do you mean? It took about 20 seconds for ksolve to generate those cases (but I do have optimal solutions for them all now ) so at about 100 cases/sec it would take a LOT of time for anything serious like pyraminx or 2x2. I imagine an algorithm which goes through the state tree once would take a lot less time than one which separately computes each optimal solution, though.


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## mrCage (May 5, 2010)

qqwref said:


> Hm, what do you mean? It took about 20 seconds for ksolve to generate those cases (but I do have optimal solutions for them all now ) so at about 100 cases/sec it would take a LOT of time for anything serious like pyraminx or 2x2. I imagine an algorithm which goes through the state tree once would take a lot less time than one which separately computes each optimal solution, though.


 
I remember the ridiculousslow times of ksolve for some domino cube cases when the optimal solution was more than 10 turns. Maybe i had bad definition file or used old ksolve version??

PS! How would you do a pure U-layer supertwist?? (UFR+,UFL-,UBL+,UBR-)

Per


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## coinman (May 5, 2010)

For the case in the end of this video i think the jumbling move is really useful http://www.youtube.com/watch?v=9RyBURilLzY&feature=related


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## coinman (May 7, 2010)

Has anybody fund a good alg for flipping two corners? I found one but it's 24 moves, i like something shorter. 
Right now i'm trying to find algs for al corner cases so i can solve F2L then al centers and finally the four remaining corners, this method might be worth trying. 

I also found a short alg that swapped two corners without turning them. I think it was les then ten moves + a double jumble to swap four centers back as seen in the video above, but i totally forgot what i did


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## qqwref (May 7, 2010)

There's [DF, FR UR UB UL UB UR UF FR], 18 moves, but it's still pretty long. I seriously think you are going to have better luck (from an efficiency standpoint) doing the corners before the centers.


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## mrCage (May 7, 2010)

qqwref said:


> There's [DF, FR UR UB UL UB UR UF FR], 18 moves, but it's still pretty long. I seriously think you are going to have better luck (from an efficiency standpoint) doing the corners before the centers.


 
Helicopter solving is still in its infancy. Lets not kill off any ideas prematurely. I have alredy posted an alg for twisting 2 corners:

[FR FD FR , FU] [FD FR FD , FL] (16 turns)

Skip the first turn (cyclical shift) for twisting 2 diagonally opposite cornera inatead, also 16 turns. Still working on the super twist. I'm down to 22 turns, due to partial cancellation (still twicw the 12 turn 3- twist)

Per


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## Portponky (May 7, 2010)

You can solve all the last layer corners in 16 turns with the method I posted so unless you are making big savings on the centers this doesn't seem worth it. What is your method for the last layer centers?


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## mrCage (May 7, 2010)

Portponky said:


> You can solve all the last layer corners in 16 turns with the method I posted so unless you are making big savings on the centers this doesn't seem worth it. What is your method for the last layer centers?


 
I am talking about twisting 4 corners purely, with centers unaffected. Personally i do corners first then centers. I am not fast at all, nor is that a priority  I guess i need 5-10 mins for each solve. I willwork more on some pretty patterns as time goes by ...

Per


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## coinman (May 7, 2010)

mrCage said:


> [FR FD FR , FU] [FD FR FD , FL] (16 turns)
> 
> Skip the first turn (cyclical shift) for twisting 2 diagonally opposite cornera inatead, also 16 turns. Still working on the super twist. I'm down to 22 turns, due to partial cancellation (still twicw the 12 turn 3- twist)
> 
> Per



After posting the above i also found a 16 move corner flip (maybe optimal?). It's basically a combination of a 3-cycle Per came up with and one i cam up with my self. But after studding the two 3-cycles i see they are the same from different angels, as is the whole corner flip alg the same Per came up with but in U instead of F 
Pers 3-cycle (RU FU RU BU)*2 my 3-cycle (FU RU FU LU)*2 .

I also now rediscovered how to swap two corners without turning them. It takes a four centers jumbling twist. Just do UR UL jumble UF UR UL. You can save a quarter turn on the first UL by stopping in the jumbling position. 



Portponky said:


> You can solve all the last layer corners in 16 turns with the method I posted so unless you are making big savings on the centers this doesn't seem worth it. What is your method for the last layer centers?



I didn't study the method you got to solve the LL corners in 16 moves but maybe i should, is it hard to recognize the cases? I solve the corners intuitionally, this can take quit a lot of moves. 
On the LL centers i mainly used the (RU FU)*3 and the (UF FR UR FR)*2 sometimes i also use (UR UF UL)*4 that moves six centers in two three cycles and the double jumble to swap four centers

The problem i se with solving the centers last is that it varies a lot from solve to solve how many centers you have to move. A lot of times they are also spread out in ways that force you to do set-up moves to solve them with the algs above. Or is there ways to control this?

Solving the centers after F2L is easy if you don't have to preserve corners so i was thinking that a method for LL corners that don't effects centers cud be more efficient.

My best time with centers last is so far 1:56.13 and after only a few solves 2:14.05 with corners last.

Edit: Now after only five-six solves with corners last i beat my PB, 1:43.47


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## qqwref (May 7, 2010)

Hi, I computed some optimal 5-gen (allowing UF, UR, UB, UL, FR) edge twist cases. I tried all 4 orientations but allowing FR didn't improve the optimal solution at all so I guess using LL twists only is fastest. These are all LL-only optimal solutions I got, up to y rotations.

"T" orientation (adj. 2 twist):
- [UR (UF UB) UL (UF UB)]2 (12f*)
"U" orientation (adj. 2 twist, inverse of T):
- [(UF UB) UL (UF UB) UR]2 (12f*)

"L" orientation (opp. 2 twist):
- [(UF UR UF) (UB UL UB)]2 (12f*)
- [(UR UF UR) (UL UB UL)]2 (12f*)

"sune" orientation (3 twist):
- [UF UR UB UL]3 (12f*)
"antisune" orientation (3 twist):
- [UF UL UB UR]3 (12f*)

"H" orientation (symmetrical 4 twist):
- [(UF UB) UR UB (UR UL) UF UR]2 (16f*)
- [(UF UB) UR UF (UR UL) UB UR]2 (16f*)
- [UF (UL UB UL) (UR UB UR UB)]2 (16f*)
- [(UF UL UF) UB (UR UF UR) UL]2 (16f*)
- [UF (UR UB UR) (UL UB UL UB)]2 (16f*)
- [(UF UR UF) UB (UL UF UL) UR]2 (16f*)
- [(UL UB UL UB) (UF UL UF) UR]2 (16f*)
- [(UL UF UL UF) (UB UL UB) UR]2 (16f*)

"Pi" orientation (asymmetrical 4 twist): 
- [(UB UR UB) (UF UR UF) (UR UL)]2 (16f*)
- [UB UR (UL UF UL UF) UB UL]2 (16f*)
- [(UF UR UF) (UB UR UB) (UR UL)]2 (16f*)
- [UF UR (UL UB UL UB) UF UL]2 (16f*)
- [(UR UB UL UF) (UB UR UB) UL]2 (16f*)
- [(UR UB UR) (UF UB) (UL UB UL)]2 (16f*)
- [UR UF (UB UR UB UR) UL UF]2 (16f*)
- [UR UB (UF UR UF UR) UL UB]2 (16f*)
- [(UR UF UL UB) (UF UR UF) UL]2 (16f*)
- [(UR UF UR) (UF UB) (UL UF UL)]2 (16f*)
- [(UR UL) (UB UL UB) (UF UL UF)]2 (16f*)
- [(UR UL) (UF UL UF) (UB UL UB)]2 (16f*)




coinman said:


> I also now rediscovered how to swap two corners without turning them. It takes a four centers jumbling twist. Just do UR UL jumble UF UR UL. You can save a quarter turn on the first UL by stopping in the jumbling position.



Ah, of course. This is a quite nice way to do this. I don't mind that it involves jumbling because you'd have to jumble to get to this position anyway


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## Portponky (May 7, 2010)

coinman said:


> I didn't study the method you got to solve the LL corners in 16 moves but maybe i should, is it hard to recognize the cases? I solve the corners intuitionally, this can take quit a lot of moves.



It's really easy to orient corners optimally, the pattern is the same as the sune move.

Permutation is pretty straight forward. You can get very good result by spamming these three moves:

U{LR B LR} - swap the two corners on UF
U{LR BF LR} - swap the two corners on UF and the two on UB
U{RBR LFL RBR} - diagonal swap UBL and UFR corners



coinman said:


> On the LL centers i mainly used the (RU FU)*3 and the (UF FR UR FR)*2 sometimes i also use (UR UF UL)*4 that moves six centers in two three cycles and the double jumble to swap four centers
> 
> The problem i se with solving the centers last is that it varies a lot from solve to solve how many centers you have to move. A lot of times they are also spread out in ways that force you to do set-up moves to solve them with the algs above. Or is there ways to control this?



I use the same method and the setup moves and bad corner distribution can be a pain. I think this will improve by figuring out / learning more center permutations.



coinman said:


> Solving the centers after F2L is easy if you don't have to preserve corners so i was thinking that a method for LL corners that don't effects centers cud be more efficient.



I just had a shot at this and you are right, the centers are quite easy. If there were full corner orient/perm algs it would be fast, but one of the major problems is you can't rotate the top layer so you get four times as many cases as a face turning cube.


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## coinman (May 7, 2010)

qqwref said:


> coinman said:
> 
> 
> > I also now rediscovered how to swap two corners without turning them. It takes a four centers jumbling twist. Just do UR UL jumble UF UR UL. You can save a quarter turn on the first UL by stopping in the jumbling position.
> ...



Na you can get to this position without jumbling but the only alg i know is looong. Do FU RU FU LU FU RU FU now four centers and two corners to the left are swapped. The four centers you can fix by turning the cube so you have them in FR and do ((UR FR)*3 (UF FR)*3)*3 = 25 moves so i prefer doing it the jumbling way 



Portponky said:


> U{LR B LR} - swap the two corners on UF
> U{LR BF LR} - swap the two corners on UF and the two on UB
> U{RBR LFL RBR} - diagonal swap UBL and UFR corners



I already found the two first algs but not the "U{RBR LFL RBR} - diagonal swap UBL and UFR corners". But trying this i wanted to restore the centers and by adding one more UL UF UL i found a 12 move diagonal corner twist. The shortest one this far, and i think optimal this time. The whole alg being U{RBR LFL RBR LFL}
And if you start with UL and finish with UF it will twist two parallel corners in the front in 12 moves, nice! The whole alg being U{LRB LRF LRB LRF}.


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## qqwref (May 7, 2010)

coinman said:


> qqwref said:
> 
> 
> > coinman said:
> ...


Oh, hang on, I misread you and thought you were talking about swapping two *centers* without turning them. Doing UL UR and then the jumbly UF swap, then UF UL UR, will swap two U centers and two others, thus quickly solving the parity issue which can come up when solving a jumbled cube.

I can switch two corners (with normal orientation) and leave centers alone in 15 moves, but there might be something better out there.



coinman said:


> I already found the two first algs but not the "U{RBR LFL RBR} - diagonal swap UBL and UFR corners". But trying this i wanted to restore the centers and by adding one more UL UF UL i found a 12 move diagonal corner twist. The shortest one this far, and i think optimal this time. The whole alg being U{RBR LFL RBR LFL}
> And if you start with UL and finish with UF it will twist two parallel corners in the front in 12 moves, nice! The whole alg being U{LRB LRF LRB LRF}.



Nice find, but I already posted these on the last page  And yes, they're optimal.


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## coinman (May 8, 2010)

qqwref said:


> coinman said:
> 
> 
> > qqwref said:
> ...



Aha, i now se that i missed your algs. They are virtually the same as the ones i came up with but from different angels. 

This might sound a bit stupid but i newer used computer solvers, is there really a solver for helicopter cubes? Or is this Ksolve a general solver for al types of cubes? If so, how do you get it to work for something as different as a helicopter cube? 

I like the try and error way, but i also started this way in 1980  Computers take away lot of the fun in exploring the cubes i think, but i guess it's useful when noting else works or the algs you find manually are way to long.


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## qqwref (May 8, 2010)

coinman said:


> This might sound a bit stupid but i ne*v*er used computer solvers, is there really a solver for helicopter cubes? Or is this Ksolve a general solver for al*l* types of cubes? If so, how do you get it to work for something as different as a helicopter cube?



It's a general solver for pretty much anything (there is no dedicated Helicopter Cube solver). You have to define the available pieces and what each available move does to them; then you can make a scramble file and it will search for sequences. I think it comes with an explanation of how to use it.
File is at http://www.svekub.se/files/ksolve.zip if you want to try.

Finding algs by hand can be fun, but when questions come up like "what's the optimal way to do this" I prefer to just go straight to computer solvers if possible. I'd rather have a solution I know is optimal than have a good alg that I can't improve, anyway.


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## mrCage (May 8, 2010)

qqwref said:


> coinman said:
> 
> 
> > This might sound a bit stupid but i ne*v*er used computer solvers, is there really a solver for helicopter cubes? Or is this Ksolve a general solver for al*l* types of cubes? If so, how do you get it to work for something as different as a helicopter cube?
> ...


 
I prefer algs that are either easy to remember and/or easy to understand. Optimality is nice too but not really my primary goal

PS! Check the 333 pattern request thread. I prefer my 26 turn way to do this. i dont have have to memorise. I cycle the 3 bars in 14 turns and then twist 2 corners in 12 turns. "Only" 9 turns from optimal 

Per


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## mrCage (May 8, 2010)

Portponky said:


> You can solve all the last layer corners in 16 turns with the method I posted so unless you are making big savings on the centers this doesn't seem worth it. What is your method for the last layer centers?


 
Well i can do the centers in 2 steps. Position then orient - PCOC.

First step takes only 2 turns, while second step takes only 12 turns maximum. There may be cancellations. I will be back with details for second step.

Per


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## mrCage (May 8, 2010)

Portponky said:


> You can solve all the last layer corners in 16 turns with the method I posted so unless you are making big savings on the centers this doesn't seem worth it. What is your method for the last layer centers?


 
Well i can do the corners in 2 steps. Position then orient - PCOC.

First step takes only 2 turns, while second step takes only 12 turns maximum. There may be cancellations. I will be back with details for second step.

Per

OK, brief details:

2-twist: UR UF UR UB UR UF UR UL or UR UB UR UF UR UL UR UF
3-twist: (UR UB UL UF)*3 or inverse
4-twist: (UR UL UB UF)*2 or UL UB UF UR UL UB UB UR


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## qqwref (May 8, 2010)

mrCage said:


> I prefer algs that are either easy to remember and/or easy to understand. Optimality is nice too but not really my primary goal


You didn't look at the algs I posted for optimal pure flips - they are all easy to remember since every single one is something repeated 2 or 3 times. (For now I treat this as a coincidence but I'd like to understand why later.) I don't think you should shun optimal algs because in the case of the Helicopter Cube each turn does pretty obvious stuff so it is easy to understand how almost any alg works. I don't know of any way to do more than one useful turn at once so optimal algs are always faster as well. Fewer moves in this case should be easier to understand, not harder.



mrCage said:


> Well i can do the corners in 2 steps. Position then orient - PCOC.
> 
> 2-twist: UR UF UR UB UR UF UR UL or UR UB UR UF UR UL UR UF
> 3-twist: (UR UB UL UF)*3 or inverse
> 4-twist: (UR UL UB UF)*2 or UL UB UF UR UL UB UB UR


Interesting, this may be closer to optimal than the other solution. Improved algs for 2-twist (I did _not_ use computer for these, just my knowledge):
2-twist: UF UR UB UR UF UL (or inverse) or UF UR UF UB UL UB

The permutation step takes 0 moves for solved, 1 for J, 2 for E, 3 for hourglass, 4 for U2, 3 for U', 3 for Y, 2 for A (I hope you can figure out what I mean for all these ). So a maximum of 4 moves, then your solution has a maximum of 16.

For the interested my style is essentially solve one, solve another, solve last two. I don't know how efficient this is. The worst case for last 2 is 9 moves.


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## mrCage (May 8, 2010)

qqwref said:


> mrCage said:
> 
> 
> > I prefer algs that are either easy to remember and/or easy to understand. Optimality is nice too but not really my primary goal
> ...


 
No, my 2 step corners has a maximum of 14 turns. I forgot to account for some PC cases that require 3 or 4 turns - but due to symmetry of these cases, we always get good cancellations combined with step 2.

Per


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## Kenneth (May 8, 2010)

Ok, from now we use scrambling, chickens!!

CO, and you think that is the same as corner? orientation??



Spoiler









Aha, Centre! orientation, that's something!!


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## Portponky (May 8, 2010)

Is there a way of jumbling it out of cube shape so that it remains stable? Usually it just seems to lock up the jumbled bit.


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## Kenneth (May 8, 2010)

The state I have you do easy using a 4 move 2 gen alg (RF) and is compleatly stable.

UR 72+ UF 180 UR 180 UF 72+

the numbers are the number of degrees you shall turn the side (3x3 R = R 90+)

Edit : I think we can say there are levels on this puzzle, you stay in cube shape, with or without jumbling (the kind that just temporarly shape shifts) or else use full shape shift... that is far much harder to solve, mosty because you get confused from the odd looks of it.

Edit more: from that I got this great alg: R72+ F180 R180 F180 R180 F180 R108+ ... double 2 cycle centres.


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## coinman (May 8, 2010)

This is my new V-cube


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## Portponky (May 9, 2010)

Kenneth said:


> The state I have you do easy using a 4 move 2 gen alg (RF) and is compleatly stable.
> 
> UR 72+ UF 180 UR 180 UF 72+



This does lock everything up. After doing it you can only turn: UF, BR, BL, DB, DL. The only edge with jumbled pieces you can turn is UF so it's fairly obvious you have to backtrack through that.



Kenneth said:


> Edit more: from that I got this great alg: R72+ F180 R180 F180 R180 F180 R108+ ... double 2 cycle centres.



If I understand your notation right, this doesn't actually leave it jumbled and from the resulting state it can be resolved with UF UR UF UR UF UR FR UF UR FR UF FR UF FR UF UR UF FR


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## Kenneth (May 9, 2010)

Portponky said:


> Kenneth said:
> 
> 
> > The state I have you do easy using a 4 move 2 gen alg (RF) and is compleatly stable.
> ...



That was just an example, I have gotten it into states where you can turn more sides. I was just now in a situation that took me about an hour to resolve to cube shape. I even found a jumble move that is doable with just a little force and it left two oriented centres when i got it back to cube shape (and possibly one oriented corner!! but it may happen when I poped it to to fix the centres). Don't now if it shall be considered a valid move though.

I did not say that 2+2 centres swap was unsolveable using normal moves.. but it is a short and fairly easy to do alg.

Actually I think it is possible to solve the whole puzzle in a 72 degree angle from where it is from start because I had a square of centres on one side and another on a adjacent side that was angled 72 degrees from the first.

Jaap, if you see this, never ever you will be able to calculate the number of possible states for this one


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## Kenneth (May 9, 2010)

Made a little video showing some shapes shiftings:



Spoiler











Edit, took me 10 minutes to get back from there, in the end I had a tricky situation with 2+2 oriented centres sitting in diagonal (UF and BD sides) with a number of moves possible, then I found a solution for that state too =)


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## mrCage (May 9, 2010)

Kenneth said:


> Made a little video showing some shapes shiftings:
> 
> 
> 
> ...


 
Hi Kenneth. Care to make a slow motion video of a jumbling sequence (from solved) non-solvable with normal non-jumbling turns!! Please!!!

Per

PS! I did appreciate that 7 turn jumbling double swap.


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## qqwref (May 9, 2010)

Kenneth said:


> Jaap, if you see this, never ever you will be able to calculate the number of possible states for this one



Including jumbling, you mean, right? I think it could be calculated, it won't be easy, but it could be done. Of course we have to define what a state is carefully. Do we want to count (for instance) solved state and solved state with UF+72 as the same, or are they different because different moves can be done in each position? How much of the puzzle can actually be jumbled? Does it still count as a state if many axes can't be turned? And so on. There is a clear limit to how much jumbling can take place (each axis holds two groups of pieces, and each turnable group MUST hold two centers and a corner).

Here is an interesting jumble: (UB- RD- FL-) (FD- RB- UL-). It leaves three 'twisted' center pairs and only three turnable axes.

Interestingly there are over 24000 times as many cube-shape states possible if you allow jumbling than possible if you don't. Because a jumbling move can typically only move 2 centers out of place at once (of 24), it would require a huge number of moves to evenly, randomly scramble among these states!



mrCage said:


> Care to make a slow motion video of a jumbling sequence (from solved) non-solvable with normal non-jumbling turns!! Please!!!.



Try this: UL- FR- UF180 UL+ FR+ UF180.


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## Portponky (May 9, 2010)

I doubt you can solve it at a 72 degree angle because the core doesn't rotate when you jumble it.

One of the fun jumbling things to do is turn UR, FL and DB all to an angle and then jumble each intersection by ninety degrees. It looks like a big mess.


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## Kenneth (May 9, 2010)

Per, yep, I can... but a little later, have some other things atm.

Michael, yes complicated, but with a clear definiton as you say, then mabye.

Portponky, ok, mabye not 72, but... we will see =)


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## coinman (May 9, 2010)

Michael, can you try to find A-perm algs (Clockwise and anti-clockwise.) with the Ksolver? I don't know how to use it 
Preferably algs that preserves the UFR corner.


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## qqwref (May 9, 2010)

There are two optimal algs (for counterclockwise perm):
UF UR (UL UF UL UB)2 UR UF (12f*)
UR UF (UL UB UR UB)2 UF UR (12f*)
You can just invert it for the clockwise one.


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## coinman (May 9, 2010)

I have now published a embryo to a heli cube corners last guide http://www.speedsolving.com/forum/showthread.php?p=377177#post377177

Pleas comment, advise and give me more algs!

All calculations of numbers of cases is done by my brother so don't blame if it's wrong


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## Kenneth (May 9, 2010)

Kenneth said:


> Portponky, ok, mabye not 72, but... we will see =)



Not! I was in a hurry and just read briefly what you wrote, now I looked at it and yes, you are right, it is not possible.

BTW guys, I just got it and have no energy to read the whole tread, I got a situation where I can solve all byt two swapped centres, did that happen to you too?

The method I tried so far is mostly intuitive, solve D face centres, solve rest of first half centres, 7 mover to put D corners, solve L and B LL side centres, use nx(UR UF) to solve the rest of the centres then "Niklas" + some algs my brother showed me for LL corners.

7 mover, to solve RFD, oriented in U at ULF, do RF UR UB UL UB UR RF (or the mirror setup and solving). It is intuitive, you move the two centres from RFD to top half, then split them, move in the corner and back up.

With no jumbling and that method it feels much like solving magic exept for the last corners 

So now I'm trying jumbling scrambling and jumbling solving using the same sub steps.


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## Portponky (May 9, 2010)

Kenneth said:


> BTW guys, I just got it and have no energy to read the whole tread, I got a situation where I can solve all byt two swapped centres, did that happen to you too?



Yes, there is a parity situation caused by jumbling. My rough unjumbling guide has an algorithm for solving this. It's actually pretty simple, any jumble with matching colours top and bottom will fix the parity case.

It's really hard to notate the jumble moves :confused:


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## coinman (May 9, 2010)

Kenneth said:


> BTW guys, I just got it and have no energy to read the whole tread, I got a situation where I can solve all but two swapped centres, did that happen to you too?



Yes!
But i have a alg that swaps two centers only. UR UL jumble front left or right corner UF UR UL. 

This thread is also about patterns but i haven't seen any 
I found a nice star and did it on two sides of the cube. It takes no jumbel moves to do it. 





The other side.


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## mrCage (May 10, 2010)

qqwref said:


> Kenneth said:
> 
> 
> > Jaap, if you see this, never ever you will be able to calculate the number of possible states for this one
> ...


 
Hmmm. Again adouble swap. Doable without jumbling. :confused:

Per

PS! My bad. Doable, but i wanted a single swap. Doable too, right!!??


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## jaap (May 10, 2010)

qqwref said:


> Kenneth said:
> 
> 
> > Jaap, if you see this, never ever you will be able to calculate the number of possible states for this one
> ...



Right. For simplicity I would consider each of the three ways to align an 'axis' with its adjacent ones to be a valid position, regardless of whether the 6 pieces actually around that axis allow any of the adjacent ones to turn. This analysis wouldn't be very hard. I might try this out soon, while I wait for this puzzle to be delivered to me.

However, this only takes into account whether something can or cannot move due to the presence or absence of the cut. It does not take into account interference with the movement due to the exterior shape of the pieces. Even pieces that are kind of non-adjacent interfere, for example the state of the UL axis can block the UR axis from turning completely freely. I don't know how much of a restriction this is, or if this causes there to be shapes that you could assemble but not reach by twisting unless you filed the pieces down.



mrCage said:


> qqwref said:
> 
> 
> > mrCage said:
> ...



No it isn't. We already established that without jumbling centres lie in 4 orbits of 6 pieces. This double swap exchanges pieces from different orbits. A double swap is only possible without jumbling if each swap lies within one orbit (though the two swaps need not be in the same one).



mrCage said:


> My bad. Doable, but i wanted a single swap. Doable too, right!!??



A single swap of centres is not possible I think, even with jumbling. The restrictions qqwref describes above pretty much prove it.


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## mrCage (May 10, 2010)

jaap said:


> A single swap of centres is not possible I think, even with jumbling. The restrictions qqwref describes above pretty much prove it.


 
I managed to do this with jumbling actually. Don't ask me how Now i can't solve the puzzle. I'll try some more jumbling soon and i'll see what comes out of that.

Per


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## mrCage (May 10, 2010)

coinman said:


> Kenneth said:
> 
> 
> > BTW guys, I just got it and have no energy to read the whole tread, I got a situation where I can solve all but two swapped centres, did that happen to you too?
> ...


 
Sequence??

Per


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## mrCage (May 10, 2010)

I will soon post a supertwist pattern (8 corners and without jumbling). The solution, so far, is quite lengthy - but it uses stuff already posted here. Combination of 3 sequences ...

Per


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## jaap (May 10, 2010)

mrCage said:


> jaap said:
> 
> 
> > A single swap of centres is not possible I think, even with jumbling. The restrictions qqwref describes above pretty much prove it.
> ...



Then you'll have to do a double swap, but set it up so that one of the swaps exchanges two pieces of the same colour. If you had a superhelicopter cube (i.e. all pieces distinct) then you cannot so a single swap.


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## Portponky (May 10, 2010)

The jumble move swaps four centre pieces, and also changes the corners in a way which can be commuted back with standard moves.

It swaps one centre piece from orbit A with a centre piece from orbit B.
It also swaps two centre pieces on orbit C.
(where A, B, C are three of the four centre orbits)

So you can do a single centre swap by arranging the centre in orbit A and the centre in orbit B to be the same colour. That way two centres in orbit C will swap and you will get a single swap situation. I have an algorithm in my guide which does this without scrambling up the almost-solved cube.

It should be noted that the single swap situation is actually orbit parity.


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## Ton (May 10, 2010)

Imitative I would say

1 center, corners of that layer 
Than the all the centers , than the last four corners 

Did not figure out a method , just found a 3 cycle of edges and a 3 cycle of corners yesterday, together with the trivial 2 cycle of corners I can solve it


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## coinman (May 10, 2010)

@Per. The Sequence is: two setup moves BR UL, then do the two corner flip alg [(UL UR) UB (UL UR) UF]*2 UL BR. Now you have two flip corners across the cube. Move three centers at a time around this flipped corner with (UR FR LR FR)*2 This alg is preformed three time per corner + cube rotations. If you don't get it please let me know.


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## coinman (May 10, 2010)

Ton said:


> Did not figure out a method , just found a 3 cycle of edges and a 3 cycle of corners yesterday, together with the trivial 2 cycle of corners I can solve it



There is two different methods (so far). One is doing F2L, LL corners, and LL + side half centers last. The other method is to do F2L, LL centers + side half centers, and LL corners last. I came up with idea to do the LL corners last and i hope and believe this is more efficient. I started a guide for this method, it's not ready but already usable. http://www.speedsolving.com/forum/showthread.php?t=21065


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## Portponky (May 10, 2010)

Do you have any idea of the max/average move counts for either method? I already posted that the LL corners can be solved with very simple algs in 16 moves. From there I would guess the centres would take something like 25 average. What about centres first method?


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## coinman (May 10, 2010)

Portponky said:


> Do you have any idea of the max/average move counts for either method? I already posted that the LL corners can be solved with very simple algs in 16 moves. From there I would guess the centres would take something like 25 average. What about centres first method?



I am not very good at calculating this kind of things but i think that the worst case for solving the centers after F2L is about 15 moves, average maybe 10. This is using the method to solve the centers after F2L mentioned in my corners last guide. Optimal is probably less, maybe 12 moves worst case? 

I think LL corners can be solved in two steps in about 20 moves average.

As an example, if i have one corner in the right position but flipped and i chose to move the rest to the right position - i think this step is max 12 moves, and then flips them all - max 16 moves, then the worst case is 28 moves. A lot of the 3 corner cycles is preformed in 8 moves and the two corner flip is 12 moves so a good solve this way is 20 moves without skips. 

The second way to do this is to solve and orientate one corner to it's final position in one alg, i think this always can be done in 8 moves and then solve the remaining 3 corners in one alg, which i think is 12 moves max. A bad solve this way is then 20 moves and a good is 16 moves (two 3 corner cycels) without skips. 

I do it the first way most of the times now sins i know to few algs to be able to always solve the remaining corners after step one in the second way.

I will ask my brother to calculate this more correctly! 
But i think i'm not to far off.

BTW i just got a 1:07.xx solve corners last but it was a bit lucky (only corner orientation after fixing centers), my non lucky PB is now 1:26.80.
OH 8:59.03


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## mrCage (May 11, 2010)

coinman said:


> @Per. The Sequence is: two setup moves BR UL, then do the two corner flip alg [(UL UR) UB (UL UR) UF]*2 UL BR. Now you have two flip corners across the cube. Move three centers at a time around this flipped corner with (UR FR LR FR)*2 This alg is preformed three time per corner + cube rotations. If you don't get it please let me know.


Why not the same pattern, but without the twisted corners? Also looks nice!!

Per


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## Kenneth (May 11, 2010)

jaap said:


> Kenneth said:
> 
> 
> > Jaap, if you see this, never ever you will be able to calculate the number of possible states for this one
> ...



That is what made me write that =)

There are so many restrictions but still so many combinations, I don't think we even found all possible moves yet and when/how do we know we have?

Creating a 3D model than checks if any piece in the actual layer has a corner under the turning plane or if any piece around it is sticking up and blocks the way and then letting it brute force for all combiantions is the only way I can think of.


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## Kenneth (May 11, 2010)

Intresting walk around jumbling with useful effects:

UR(72) LF UF LF UF(108) UR(-72) UF LF UF(-108) LF

Add UF UR UF UR 

Do first UR(-108) and it can do "2 centeres" (3 cycle but 2 U) and 2 diagonal LL corners...

If you are a bad boy you can even cheat : UR(72) UF(-72) UL(-72)!? UF!! UL(72) UF(72) UR(-72) 

Warning, it pops during the mid UF if you try counter clockwise, go CW and it works.

More madness but valid moves, it leaves two fliped centres and corner orientation parity!!!

UR(72) UF UR(108) LF(-72) UB(108) UL(72) (Nice shape here) LF(-72) UB(72)

Per, never mind that video, I tried but it was so hard to follow the moves because of shapeshifts and the fact that I constantly had to rotate it so it faced the camera, it was just confusing.


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## Kenneth (May 11, 2010)

Two seconds penalty?



Spoiler










I was trying to speedsolve, but why on earth did I miss my last U move :fp


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## coinman (May 11, 2010)

New simple pattern, the white and yellow sides are intact.


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## mrCage (May 11, 2010)

Kenneth said:


> Two seconds penalty?
> 
> 
> 
> ...


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## Kenneth (May 11, 2010)

I think we need to agree about some nice notation for the turns that are not 180 degrees. Each side have six diffrent positions. 0, 72+, 108+, 180, 108- and 72-. Would it be wrong to use UR1, UR2, UR3 (alt UR only as is used already), UR2' and UR1' ? 

Note that this is as for 3x3, not fixed positions but how much you shall rotate relative to the posittion the side has when you start the turn.

Some terms might be nice too, I think "F2L" is wrong, FLC = first layer cubies, LLC, last layer cubies? 

More?


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## qqwref (May 11, 2010)

Of course F2L is wrong, it's just first layer and last layer. There are pretty clearly two layers on this thing. If you want to talk about just the centers (or corners) on one side you can talk about sides (so FS centers would be the first step of my method).

I like the idea to use six positions no matter what (to describe jumbling) but maybe it would be better to use UR+, UR2+, UR3+, UR2-, UR- or something like that, just to emphasize that we aren't talking about normal turns. Putting something like UR3 might be confused with just doing UR 3 times.


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## Kenneth (May 12, 2010)

Oki, FL and LL then =)

For notation, I use your style. To expand it a bit, I got this idea last night: many algs are having moves like UR UL xx UR UL yy ... and so on, to shorten these why not use H = UR + UL and I = UF + UB?

Then a H-PLL is H I H I H I or 3X( H I )

Or even UH and UI, then FH = FR FL and FI = UF DF

Then the H-PLL is 3x( UH UI )

A bit like slice notation... mabye a special metric?

-------------------------

I'm looking a bit at CLL, if I get it right the whole group of 192 cases may be reduced to 80 cases; 8 orientations (solved, U, T, L, S, -S, H and pi) times 10 permutations (solved, A, A', H, E, O, O', M, diag and adj two swap). Then there are some duplicates too but I have not fully checked all of them.

But if you solve one corner (1+11 cases) it will be a group of 54 left that is reduceable to 44 (18 3-cycles, 18 2-swap + 1 CO, four pure CO and four cases with 2 corners solved) and solving one corner only is almost intuitive, if you did not jumble it you can do that before the last two adjacent centres (that always is 2-gen solving).

Nice CLL using walk around jumbling:

UR1+ UF UR2+ UB UF UR2+ UB UR1+ (UF UB UR UF UB UR).

OR with the "slice notation" : UR1+ UF UR2+ UI UR2+ UB UR1+ (UI UR UI UR).

But, it is not possible to do the first UI in any order =)

BTW, I timed a fully scrambled solve last night and it was 8:57 minutes, (cube shape was compleated around 1:40 and then many errors during FL) I claim this to be the UWR until someone claims a better one 

I used a stopwatch but to my suprise it was stackmatable...

EDIT: hmm, maybe COLL = preserve LL centres (and FL of course) and CLL = ignore LL centres? ... then it is COLL I'm looking at =)

-------
I found a few PLL's too, but how useful is that? half a billion cases and horrible recognition


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## ribonzz (May 12, 2010)

Helicopter cube eh? I just want to order it.


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## coinman (May 12, 2010)

Did my first avg of 12 today, result 1:45.88. and a new non lucky single PB 1:15.55. 
Messed up a few solves so i think i can do better. Still need to learn more algs.


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## qqwref (May 12, 2010)

Kenneth said:


> BTW, I timed a fully scrambled solve last night and it was 8:57 minutes, (cube shape was compleated around 1:40 and then many errors during FL) I claim this to be the UWR until someone claims a better one


Clearly I could beat this, but... how do you define "fully scrambled"? If you provide a scramble I will happily time it for you. PS: qqTimer will scramble helicube completely without jumbling, so you can just add turns onto that.


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## Kenneth (May 12, 2010)

I'm sure I also can beat that, it was my first ever attempt to time a full solve.

When scrambling I did shape shift from the first move and then jumbled around for a while. When I was in a position where two adjacent sides was possible to turn I did a random number of 2 gen moves on them.

Locks may be unlocked by turning one of the locking sides, like in this example

UR1+ UF UR2+ -- here it is not possible to turn UF but if you do UB it is. Keep doing like so until you got a good scramble.

I can make up a long written scramble later, got to fix some food first...


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## Portponky (May 12, 2010)

I don't think it would be normal to scramble the puzzle in that way. This isn't really a normal shapeshifting puzzle because most of the shapeshifting moves lock the entire thing up. This means you get a really sporadic difficulty. Also it's really hard to notate what has been done.

I think that it should either be scrambled with regular moves (1st type), or jumbled using the standard jumble move that doesn't change its shape (2nd type). That seems to be more what the intention of the puzzle is.


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## qqwref (May 12, 2010)

Portponky said:


> I don't think it would be normal to scramble the puzzle in that way. This isn't really a normal shapeshifting puzzle because most of the shapeshifting moves lock the entire thing up. This means you get a really sporadic difficulty. Also it's really hard to notate what has been done.


Exactly, that is why I was skeptical of saying "this is the UWR for a fully scrambled puzzle", because it's too hard to define. Helicopter cube shapeshifting will be like the Square-1 shape, some are very easy and some take a lot of moves unless you really know what you're doing.



Portponky said:


> I think that it should either be scrambled with regular moves (1st type), or jumbled using the standard jumble move that doesn't change its shape (2nd type). That seems to be more what the intention of the puzzle is.


Yeah. I already have a scrambler of the 1st type; the 2nd type might be doable for automated scrambling, but the problem is that you'd need a lot of jumbling moves to really randomize the orbits completely - maybe 8 or more separate (FR- UL- UF UL+ FR+) type things. I don't really know if doing a bunch of those (how many?) would make a good scramble.


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## Kenneth (May 12, 2010)

Michael, try this one: 

UR1+ UF UR2+ UB BL UF1+ RF UF1- DR UR UF UR1- UB1+ FR1+ UR FR UB UR UB1- BL UB DB DF2+ UR1+ UL1- BL UL BL

It probably have to be longer because it leaves some blocks. But as a start it will do... I guess most of the blocks will be destroyed when the cube shape is getting fixed again


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## mrCage (May 12, 2010)

A while ago i promised a supertwist pattern. Well, here it is. Not all that short.

RB (setup)
(UF UR UB UL)*3 (3 twist)
(DF DR DB DL)*3 (another 3-twist)
RB (undo setup)

(FU FR FU FL)*2 (FR FU FR FD)*2 (final 2-twist)

Did not try insertions 

Per

PS! Soon to come, cube-in-cube (yes with jumbling!)


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## qqwref (May 12, 2010)

Kenneth said:


> Michael, try this one:
> 
> UR1+ UF UR2+ UB BL UF1+ RF UF1- DR UR UF UR1- UB1+ FR1+ UR FR UB UR UB1- BL UB DB DF2+ UR1+ UL1- BL UL BL
> 
> It probably have to be longer because it leaves some blocks. But as a start it will do... I guess most of the blocks will be destroyed when the cube shape is getting fixed again



Fun scramble  Solved it in 3:03.85.



mrCage said:


> A while ago i promised a supertwist pattern. Well, here it is. Not all that short.
> 
> RB (setup)
> (UF UR UB UL)*3 (3 twist)
> ...


That's 42 moves :| The top and bottom are each a symmetrical 4-twist, and I posted a 16-move alg for that before, so I already have a 32-move solution for this. Sorry


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## Kenneth (May 12, 2010)

qqwref said:


> Kenneth said:
> 
> 
> > Michael, try this one:
> ...



Nice, that cuts like 2:3 of the previous record 

It had a fairly easy shape, I solved it in 30 seconds or so, but the rest was slow...

There is a situation that has two pairs of unoriented centres sitting in "half diagonal" (UR FL for example), the rest is a cube and that I have real problems with. It is not so intuitive, I struggled for an easy 20 minutes trying to solve it earlier today, eventually I got our of it but I'm not sure what I did =)

That state is near where you can do cube in cube but having it twisted in position! No, not "stickers twisted", the physical block is. Or rather, the smaller cubies are solved and the rest is twisted. And it is not 100%, it has one centre on each half that is wrong.

-----------------

As long as I got the last post I edit it =)

Nice jumbling that swaps two blocks and leaves the rest unaffected : UR1+ UF UR1- FL1- UL FL1+ UR1+ UF UR1-

Add UB before and after and it does PLL.

Another PLL: UR1+ UF UR2+ FL1- UB2+ UL UB2+ FL1+ UR2- UF UR1-

Both PLL's only 11 turns.


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## Portponky (May 12, 2010)

qqwref said:


> Helicopter cube shapeshifting will be like the Square-1 shape, some are very easy and some take a lot of moves unless you really know what you're doing.



A square 1 is always very simple to return to a cube shape. Square 1 locks up because it is a bandaged puzzle, not because of its shapeshifting.


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## qqwref (May 12, 2010)

Portponky said:


> qqwref said:
> 
> 
> > Helicopter cube shapeshifting will be like the Square-1 shape, some are very easy and some take a lot of moves unless you really know what you're doing.
> ...



I dunno, it's not THAT simple, if you're from the perspective of someone who hasn't solved it hundreds of times... and still, some of the positions require 6 or 7 / turns to optimally solve, not even counting top and bottom turns. Sure, there are probably more (and more complicated?) shapes that a Helicopter Cube can go into, but I think the comparison makes sense. It's harder to completely scramble a Helicopter Cube mainly because you can't just go from one shape to others freely (more turns are blocked, in general).


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## mrCage (May 12, 2010)

qqwref said:


> I posted a 16-move alg for that before, so I already have a 32-move solution for this. Sorry


 
Could you repost that 16-mover please. Missed it, and the thread is long and confusing

Per


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## qqwref (May 12, 2010)

It was at the very end of the first page (4th page for you? I have 40 posts per page) where I posted pure optimal CO algorithms for the last layer:



qqwref said:


> "H" orientation (symmetrical 4 twist):
> - [(UF UB) UR UB (UR UL) UF UR]2 (16f*)
> - [(UF UB) UR UF (UR UL) UB UR]2 (16f*)
> - [UF (UL UB UL) (UR UB UR UB)]2 (16f*)
> ...


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## coinman (May 12, 2010)

Also try UR UL (x)*12 
Almost superflip in 24 moves, not counting cube rotations..

Just got a 1:03.61 lucky solve, only corner orientation after fixing centers.


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## mrCage (May 13, 2010)

mrCage said:


> A while ago i promised a supertwist pattern. Well, here it is. Not all that short.
> 
> RB (setup)
> (UF UR UB UL)*3 (3 twist)
> ...


 
Oops. So far i have needed 7 algs for the normal cube-in-cube. I will post if i find something better. Maybe a different cube-in-cube (2-cycle) across the UFR-DLB axis??

Per


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## Kenneth (May 14, 2010)

Kenneth said:


> qqwref said:
> 
> 
> > Fun scramble  Solved it in 3:03.85.
> ...



3:32.59

But I had a bit of luck in FL after the cube shape with more pieces solved than you can expect from random distrubution, so I guess my scrambling wasn't enough, more turns next time...

-----------------------------

Theory: OLL/PLL, how many are there?

OLL is to complicated for me to calculate but I think PLL is 24 CP times 8! centres (but you may have 1-4 two cycles of two centres of the same colour that will reduce the number, don't know how much though)

24 x 8! = ~1,000,000

A PLL skip in a million??


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## mrCage (May 14, 2010)

coinman said:


> I also now rediscovered how to swap two corners without turning them. It takes a four centers jumbling twist. Just do UR UL jumble UF UR UL. You can save a quarter turn on the first UL by stopping in the jumbling position.


 
Why do jumbling for that? I do it like this:

UR UF FR (UR UF)*3 FR UF UR and then FR

Per


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## qqwref (May 14, 2010)

I don't know if this is what he meant, but you can do pure swap of two corners (keeping orientation on both) with
UR UL (FR- UL- UF FR+ UL+) (UR+ FL+ UF UR- FL-) UF UR UL
Despite the relatively large number of moves the two sections in parentheses are quite fast. One way without jumbling is:
UR UL UF UL FL UF (UL FL)3 UF FL UR


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## mrCage (May 14, 2010)

qqwref said:


> I don't know if this is what he meant, but you can do pure swap of two corners (keeping orientation on both) with
> UR UL (FR- UL- UF FR+ UL+) (UR+ FL+ UF UR- FL-) UF UR UL
> Despite the relatively large number of moves the two sections in parentheses are quite fast. One way without jumbling is:
> UR UL UF UL FL UF (UL FL)3 UF FL UR


 
A pure corner swap is a swap of order 2 (swap*2=I). You restrict pure to the sense of PLL (or PC if you like).

Per


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## qqwref (May 14, 2010)

What are you talking about? Those sequences ARE pure - no centers are moved, even if you had a supercube. This is possible because a single helicopter cube turn has an odd effect (2-cycle) on corners.


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## mrCage (May 14, 2010)

qqwref said:


> What are you talking about? Those sequences ARE pure - no centers are moved, even if you had a supercube. This is possible because a single helicopter cube turn has an odd effect (2-cycle) on corners.


 
Yes, but i'm claiming my swap is pure also. Otherwise pls define pure

Per

PS! Example of impure swap would be to do my swap. Then twist UFR and UFL.

PPS!! UF is an impure corner swap with centers side effect


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## qqwref (May 14, 2010)

"Pure" means "nothing else is affected". Your swap IS pure. The difference is that yours swaps two corners on the top layer but affects their orientation (with respect to the top), whereas mine swaps two corners on the top layer without affecting their orientation.


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## mrCage (May 14, 2010)

qqwref said:


> "Pure" means "nothing else is affected". Your swap IS pure. The difference is that yours swaps two corners on the top layer but affects their orientation (with respect to the top), whereas mine swaps two corners on the top layer without affecting their orientation.


 
Then we agree! Words often come in the way

Per

PS! Do you have short pure center swap?? (jumbling obviously)

Mine is 12 turns. A conjugated 8 turn jumble.


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## qqwref (May 14, 2010)

Well, there's UL UR2- FL+ UF FL- UR- UF UR UL, which coinman (I think) posted before, but it's not really pure - it also swaps two U centers.


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## mrCage (May 15, 2010)

Neat pattern: (2 big flips)

UF - UL UF UR UF UL - UR UF UL UF UR - UF

Per

PS! Can be repeated on D layer for another striking pattern.


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## qqwref (May 15, 2010)

Kenneth said:


> Theory: OLL/PLL, how many are there?
> 
> OLL is to complicated for me to calculate but I think PLL is 24 CP times 8! centres (but you may have 1-4 two cycles of two centres of the same colour that will reduce the number, don't know how much though)
> 
> ...



What do you mean "OLL/PLL"? And the centers can't move around freely, unless you're talking about a jumbled cube, but then you shouldn't use algs...


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## mrCage (May 16, 2010)

qqwref said:


> Kenneth said:
> 
> 
> > Theory: OLL/PLL, how many are there?
> ...


 
OLL/PLL does not seem good for speedsolving, especially if jumbled

Per

PS! I shortened my solution for cube-in-cube pattern. Still 5 rather lengthy algs ... Working on a different cube-in-cube of order 2. Also tricky!


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## Kenneth (May 17, 2010)

qqwref said:


> Kenneth said:
> 
> 
> > Theory: OLL/PLL, how many are there?
> ...



I'm talking about make top side white and then permute the sides of the top layer (jumbled, otherewise there are not so many combinations). I don't think it is a good way to solve it but it is still possible.

I was just thinking of it as a theoretical matter, pure curiosity


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## mrCage (May 17, 2010)

SUPERFLIP AND SUPERFLIPTWIST

I's like to add 2 more patterns taken from 3x3x3 cube. The superflip i construct from the following 2 sequences:

(UR UF)*3 (UF LF)*3 Repeat 4 times then flip last 4 "center-edges" slightly differently.

Twist all 8 corners for the superfliptwist.

Per


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## coinman (May 18, 2010)

A new pattern.





The other side.


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## mrCage (May 19, 2010)

coinman said:


> A new pattern.


 
Order 2 or 6 ?? (show the 3 other faces)

Per

PS! Nearly every pattern is possible. Every center configuration is possible by jumbling. Only corner orientation is retricted!!


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## qqwref (May 19, 2010)

mrCage said:


> PS! Nearly every pattern is possible. Every center configuration is possible by jumbling. Only corner orientation is retricted!!


Yep, I said that already on post 2 ;-) But, even without jumbling, you can move full centers around in any way you choose. I used this trick when I swapped black and orange on mine to get the normal color scheme, because I wanted to move as few stickers as possible.


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## mrCage (May 19, 2010)

qqwref said:


> mrCage said:
> 
> 
> > PS! Nearly every pattern is possible. Every center configuration is possible by jumbling. Only corner orientation is retricted!!
> ...


 
Ok. I will take it further. On a picture helicopter cube, exactly half the center permutations are possible (with jumbling). Corners will remain the same as normal helicopter cube.

Per

PS! I mean "small" centers


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## mrCage (May 20, 2010)

Arrow pattern. I make this from 3 big flips. One flip each on FL, UB and RD edges respectively.

One big flip is made like so: UF UL FL (UF UL)*3 FL UL UF

Per


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## mrCage (May 20, 2010)

Snake pattern. This is nothing but 2 easy 3-cycles on the centres.

FL BR (setup)
(UB LB LU LB)*2 (first cycle)
(FR FD FR RD)*2 (second cycle)
BR FL (undo setup)

Sorry no picture

Per


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## qqwref (May 20, 2010)

Cool patterns. You should use a normal color scheme


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## mrCage (May 21, 2010)

qqwref said:


> Cool patterns. You should use a normal color scheme


 
Lazy to resticker Maybe I could use an applet??
Most patterns that move many pieces have too long algs ... 

How do you swap 2 big centers? 2 8-turn algs??

Per


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## qqwref (May 21, 2010)

Could you use an applet? You'd have to deal with not being able to jumble.

I don't know how to swap two center groups in two 8-move algs, I just know it can be done without jumbling. To be honest I don't care about optimizing patterns on this puzzle, just about quickly/efficiently solving.


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## mrCage (May 21, 2010)

qqwref said:


> Could you use an applet? You'd have to deal with not being able to jumble.
> 
> I don't know how to swap two center groups in two 8-move algs, I just know it can be done without jumbling. To be honest I don't care about optimizing patterns on this puzzle, just about quickly/efficiently solving.


 
There are many patterns without jumbling...

I don't care about speedsolving this thing  Just like i don't really care about speedsolving any regular cube/puzzle anymore. I'm getting old i guess

I posted a way to swap 2 face centers earlier. Ok here it goes:

1. UF UR UF FR UR UF UR FR
2. UB UR UB RB UR UB UR RB
It swaps U and R full centers.

Per


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## mrCage (May 22, 2010)

Yet abother arrow pattern:

UR FL (FR UR)*3 (UF UL)*3 FL UR

Affects only U and F faces so it can be repeated on the other 4 faces ...

Per


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## qqwref (May 23, 2010)

mrCage said:


> Yet abother arrow pattern:
> 
> UR FL (FR UR)*3 (UF UL)*3 FL UR
> 
> ...



Do you mean:
UR FL (FR UF)3 (UF UL)3 FL UR?


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## mrCage (May 23, 2010)

qqwref said:


> mrCage said:
> 
> 
> > Yet abother arrow pattern:
> ...


 
Yes. Isn't (A B)*x and (A B)x the same??

Per

PS! Hehe._ I see my mistyping now!!_


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## coinman (May 25, 2010)

I finally sprayed my helicube with some silicon, out of some reason i didn't believe it would help but it became mush better. On one of my first tries after lubrication i got 1:00.96 witch is almost 15s better then my previous pb.


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## qqwref (May 25, 2010)

Good job, soon that will be average for you


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## MichaelP. (May 28, 2010)

Where do you learn how to solve one? I just got mine in the mail today. I've solved it up to the point where (if you hold it like a normal 3x3 cube) the front face's top two center pieces need to be swapped with the bottom center closer to you on the left or right (the front center on the left needs to be swapped with the left center nearest you, and same with the rights). In this diagram the letters depict center pieces and the top block is the F face.....................................................................................................................................................................................
..........................................................................................................................................................................................
......................B G...............................................................................................................................................................
......................R R...............................................................................................................................................................
................B B........G G
................B R........R G


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## Kenneth (May 29, 2010)

MichaelP. said:


> Where do you learn how to solve one? I just got mine in the mail today. I've solved it up to the point where (if you hold it like a normal 3x3 cube) the front face's top two center pieces need to be swapped with the bottom center closer to you on the left or right (the front center on the left needs to be swapped with the left center nearest you, and same with the rights). In this diagram the letters depict center pieces and the top block is the F face.....................................................................................................................................................................................
> ..........................................................................................................................................................................................
> ......................B G...............................................................................................................................................................
> ......................R R...............................................................................................................................................................
> ...



I think it is better if you try to make a picture


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## coinman (May 29, 2010)

MichaelP. said:


> Where do you learn how to solve one?



There is two guides, my LL corners last guide: http://www.speedsolving.com/forum/showthread.php?t=21065 
And Shambles LL centers last guide: http://www.shamblessoftware.co.uk/d_puzzles_helicopter_solution

PS: My PB with my corners last method is now 57.11


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## mrCage (Jul 6, 2010)

Sorry to bump old thread. I see no need to create a new one

2 quick patterns. Pictures will follow (added!).

Pattern 1:

(FL RD UB UL FD RB)*2 

Pattern 2:

(FL RD UB UL FD RB)*3 (this can also be done like this: (UB UL)*3 (FL FD)*3 (RD RB)*3)

Per


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## mrCage (Aug 30, 2010)

Yet another pattern. Strikingly short, IMHO.

I was playing around with some trivial moves, UF UR FR. Then i simply repeated these turns 4 times. This gave the pattern with only a few missing centres. These centres were fixed with a 8 turn alg. This gave 6 cancelled turns!!:tu

Full sequence: (UF UR FR)*3 UR FR UR UF UR.

Per

PS! I also found a 16-turn sequence and a 18-turn sequence for this pattern. The 18 mover goes like so:
((UF UR FR)(UR FR UF)(FR UF UR))*2

And the 16 mover has 3 parts
1 (UF UR)*3
2 (UR FR)*3
3 FR UF FR UR UF FR UF UR

This shortens to (UF UR)*2 UF FR UR FR UR UF FR UR UF FR UF UR

Oh well, noone seems interested...

PPS! Using the last five turns of the first sequence, i found out i can simply repeat them 4 times over.
(UR FR UR UF UR)*4. This yields another 14-turn sequence...

Using (UF UR FR)(UR FR UF)(FR UF UR) from the 2nd seqence, i can insert UR UF FR (UR UF)*3 FR UF UR and then do a FR turn at the end. Another 14- turner. Ok, enough is enough!!


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## mrCage (Sep 5, 2010)

coinman said:


> Kenneth said:
> 
> 
> > BTW guys, I just got it and have no energy to read the whole tread, I got a situation where I can solve all but two swapped centres, did that happen to you too?
> ...


 
Finally i have a good sequence for this nice pattern.

(FL FD RD RB UB UL)*2 (UB UL)*2

A semi-random find:tu

Sod that. I already improved it to (FL FD RD RB UB)*2. Now beat that 

Per


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## mrCage (Sep 6, 2010)

Slight variety of the previous pattern. It consists of 4 short conjugates. The first 2 cycles 3 blocks. The last 2 cycles 3 other blocks. Both duplets twist the same 2 "center-edges' effectively cancelling the side effect.

UB UL UB . FL FD FL . RD FD RD . FL UL FL

Picture follows shortly!! (added)

Per

PS! I posted this pattern a few posts above, but a different sequence. Hmm... Pattern blindness!!


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## Portponky (Sep 8, 2010)

I really like that pattern :tu


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## mrCage (Sep 13, 2010)

Prove that my last 2 posted patterns are not possible to achieve by using only 3 different edges although all the moving pieces can be moved by 3 edges alone (rather easy!).

Prove that even 4 edges will not suffice (harder).

Are my sequences minimal?? Use for instance ksolve!!

Per


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## coinman (Feb 14, 2011)

Nice to meet you in Trondheim Per. I didn't read this thread for a while so i missed that you came up with a improved alg for my star patter. This makes it a whole lot easier to make it


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## Berd (Nov 7, 2014)

*Helicopter Cube Parity?*

So yeah, 2 'edges' need to be swapped. Anyone got an alg?


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## Dene (Nov 8, 2014)

Could you be more specific? Perhaps a picture?

I only call pieces "centres" and "corners". I'm not sure what you mean by "edges".

You might find what you need in this thread.


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## Berd (Nov 8, 2014)

Dene said:


> Could you be more specific? Perhaps a picture?
> 
> I only call pieces "centres" and "corners". I'm not sure what you mean by "edges".
> 
> You might find what you need in this thread.







So yeah, centers sorry.


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## qqwref (Nov 8, 2014)

I believe I posted a mirrored version of this alg in a recent curvy copter thread. Can't find it tho.

Anyway, one way to do it: DL DF RF (UR+ FL+ UF UR- FL- UF) RF DF DL


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## Berd (Nov 8, 2014)

qqwref said:


> I believe I posted a mirrored version of this alg in a recent curvy copter thread. Can't find it tho.
> 
> Anyway, one way to do it: DL DF RF (UR+ FL+ UF UR- FL- UF) RF DF DL



I don't quite understand the +'s and -'s in the alg.


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## Dene (Nov 8, 2014)

That's jumbling. I could never understand it by seeing it written down, and I had to have someone show it to me. But I'll give it a shot anyway. You need to take the UR and FL sides, and do a turn not quite halfway, so the corners of the triangles in the FR-middle and UL-middle meet up in the UF-middle. Then you can twist the "UF" side, which has kind of become a new piece.

EDIT: If you can make any sense of what I just said you deserve an award. Here's a video: https://www.youtube.com/watch?v=NTHQ-U3aWo0

The guy in the video does it with different sides, but hopefully you can get the idea.


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## Berd (Nov 8, 2014)

Dene said:


> That's jumbling. I could never understand it by seeing it written down, and I had to have someone show it to me. But I'll give it a shot anyway. You need to take the UR and FL sides, and do a turn not quite halfway, so the corners of the triangles in the FR-middle and UL-middle meet up in the UF-middle. Then you can twist the "UF" side, which has kind of become a new piece.
> 
> EDIT: If you can make any sense of what I just said you deserve an award. Here's a video: https://www.youtube.com/watch?v=NTHQ-U3aWo0
> 
> The guy in the video does it with different sides, but hopefully you can get the idea.



Ive spent the last 2 hours trying to solve this and I dont get it. It's like an impossible parity. Could someone help please?!


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## AndersB (Nov 8, 2014)

You could also just twist RF and resolve with an even number of RF moves


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## Dene (Nov 9, 2014)

Berd said:


> Ive spent the last 2 hours trying to solve this and I dont get it. It's like an impossible parity. Could someone help please?!



The algorithm provided by qqwref is what you need.


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## Berd (Nov 22, 2014)

Sorry for the bump but Thankyou! I solved it! Thankyou so much!


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