# Super-supercube challenge!!



## mrCage (Sep 21, 2010)

Some here may be familiar with the concept of super-supercubing. Normal supercubing involves positioning (or orienting) the center cubies (typically a picture cube).
Super-supercubing extends this concept to also involving inner shells on a larger cube. A 4x4x4 cube will also have a 2x2x2 inner cube, a 5x5x5 will have an inner 3x3x3, and so on.

Now the challenge. For a 5x5x5 super-supercube, find a way to twist 2 corners on the inner 3x3x3 without affecting the outer 5x5x5. What is the minimal length of such an algorithm? A super-super cube simulator (cubixplayer2) is available from the spedsolvingrubikscube yahoo group file section and was coded by me quite a few yrs back now

Per

PS! I'm unable to attach that file with this new design. I may find a solution later:fp


----------



## bobthegiraffemonkey (Sep 21, 2010)

Spoiler



Use a commutator like [r' d' r d r' d' r, u] (all inner slices only). r u r' u r u2 r' l' u' l u' l' u2 l also works (inner layer sune+antisune). Optimal is the same as flipping two corners on a 3x3x3. Any supercube friendly alg on a 3x3x3 will work when applied to the inner layers, since T-centers and wings follow the center orientation of the inner 3x3x3 and the X-centers follow the 3x3x3 edges. Also, circle cubes mimic these puzzles, as pieces inside the circles correspond to pieces in inner layers.


----------



## mrCage (Sep 21, 2010)

bobthegiraffemonkey said:


> Spoiler
> 
> 
> 
> Use a commutator like [r' d' r d r' d' r, u] (all inner slices only). r u r' u r u2 r' l' u' l u' l' u2 l also works (inner layer sune+antisune). Optimal is the same as flipping two corners on a 3x3x3. Any supercube friendly alg on a 3x3x3 will work when applied to the inner layers, since T-centers and wings follow the center orientation of the inner 3x3x3 and the X-centers follow the 3x3x3 edges. Also, circle cubes mimic these puzzles, as pieces inside the circles correspond to pieces in inner layers.





Spoiler



That's correct. So an optimal solution might be for instance (r' d2 r b u2 b')*2.


----------



## cmhardw (Sep 21, 2010)

I miss the super-supercube applet! I will have to download it again and fiddle with it. Yes I agree that the circle 4x4x4 cube is very similar to this concept, only it takes a regular 4x4x4 and places a 2x2x2 inside of it. If the 4x4x4 center pieces, the ones not in the circle, could be restickered to make it a supercube 4x4x4 then this would be exactly a 4x4x4 super-supercube.

Shameless plug on my website, but I am very proud of this formula. The number of combinations to a super-supercube is absolutely staggering and here is an explicit formula for it that I derived back when Per first came out with this applet.


----------



## mrCage (Sep 22, 2010)

cmhardw said:


> I miss the super-supercube applet! I will have to download it again and fiddle with it. Yes I agree that the circle 4x4x4 cube is very similar to this concept, only it takes a regular 4x4x4 and places a 2x2x2 inside of it. If the 4x4x4 center pieces, the ones not in the circle, could be restickered to make it a supercube 4x4x4 then this would be exactly a 4x4x4 super-supercube.
> 
> Shameless plug on my website, but I am very proud of this formula. The number of combinations to a super-supercube is absolutely staggering and here is an explicit formula for it that I derived back when Per first came out with this applet.


 
Chris!! Calling my windows .exe application an applet is like calling a master magic a cube:fp

Per


----------



## qqwref (Sep 22, 2010)

mrCage said:


> Spoiler
> 
> 
> 
> That's correct. So an optimal solution might be for instance (r' d2 r b u2 b')*2.


I thought of this before I saw the spoiler 

The Crazy Cube-II and III are equivalent to the 4x4 super-supercube. They also have the property of being a 2x2 inside a 4x4 but the 4x4 centers must be solved exactly.


----------



## mrCage (Sep 22, 2010)

qqwref said:


> The Crazy Cube-II and III are equivalent to the 4x4 super-supercube. They also have the property of being a 2x2 inside a 4x4 but the 4x4 centers must be solved exactly.


 
Links to these would be helpful. A quick search resulted in nothing.

Per


----------



## cmhardw (Sep 22, 2010)

mrCage said:


> Chris!! Calling my windows .exe application an applet is like calling a master magic a cube:fp
> 
> Per


 
Once again, my tech savviness shows itself haha. Wasn't done intentionally, just out of ignorance.

Chris


----------



## qqwref (Sep 22, 2010)

Crazy 4x4-II is here:
http://witeden.com/goods.php?id=36

The III is similar but the circle is slightly larger, so that it has a little piece on each corner as well. The II and III are mathematically equivalent.


----------



## mrCage (Sep 24, 2010)

New challenge(s). Now i want an edge 2-flip (adjacent edges) and a same layer 4-flip.

Per


----------



## qqwref (Sep 24, 2010)

Perhaps not optimal, but:



Spoiler



(F E F2 E2 F) U (F' E2 F2 E' F') U'
(F B E F2 E2 F B') U (F' B E2 F2 E' B' F') U'


----------



## mrCage (Sep 25, 2010)

qqwref said:


> Perhaps not optimal, but:
> 
> 
> 
> ...


 


Spoiler



Looks good. For the 4-flip consider this alg: R' D' B' D L' D2 L B D R D' F D2 F'. An alg i discovered about 20 yrs ago!!:tu
Somehow this did not work for inner cube as it left some unsolved wing centers on the outer 5x5x5. Weird! I will look into why this is so.
Consider this instead [F' R' E2 R2 E' R' F, U2]. Based on same idea as you. This was by backup solution


Per


----------



## mrCage (Sep 27, 2010)

This is the toughest one i could think of. Do the superfliptwist on the inner cube. While leaving the outer 5x5x5 intact!
The superfliptwist flips all the 12 edges and twists all the 8 corners, but such that no adjacent corners are flipped the same way

Per

PS! This thread already has enough information for making a crazy long solution! I am working on some various ideaa, but as of yet they are not very fruitful ...


----------



## mrCage (Sep 28, 2010)

Here is a rough outline of how it might be done. Make a fliptwist on 1st layer by applying the following 8 turns twice: r' d' r2 f' r' f2 d' f'. The side effect on wing centers of outer cube can be cancelled by a few outer layer conjugate turns before doing same on opposite layer. This leaves 4 edges to be flipped on inner cube on the E layer. For these i would use a conjugate version of qqwref's edge 4-flip.

How much all of this could be optimised, i have no idea ...

Per


----------



## mrCage (Oct 2, 2010)

Here is a slightly optimised complete solution. The steps are rearranged to achieve a slight cancellation before fliping the edges of the inner cube's middle layer.

1. F2 B2 R2 L2 (reposition outer cube, cage)

2. (r' u' r2 b' r' b2 u' b')*2 (inner cube top layer fliptwist)

3. F2 B2 R2 L2 (undo repositioning)

4. (r' d' r2 f' r' f2 d' f')*2 (inner cube bottom layer fliptwist)

5. [f r' d f' b u' r b', E2] (inner cube middle layer 4-flip on edges, E is the single middlemost layer obviously)

Anyone has some ideas to optimise this further??

Per


----------



## mrCage (Oct 7, 2010)

I think we have run out of flip and twist challenges. This time do a classic pattern on the inner 3x3x3: the cube in cube pattern. As always leave the outer 5x5x5 untouched!

Per


----------



## qqwref (Oct 7, 2010)

First idea:
U R' F2 R b R' F2 R b' U'
D' L B2 L' f' L B2 L' f D


----------



## mrCage (Oct 8, 2010)

qqwref said:


> First idea:
> U R' F2 R b R' F2 R b' U'
> D' L B2 L' f' L B2 L' f D


 
Hmmm. The upper case turns do nothing on the inner cube, so the first 10 moves cycles 3 edges on outer 5x5x5 and does nothing on inner 3x3x3. Similar with the last 10 moves, unless i misunderstand your notation. Hmmm.

My complete and correct solution so far has 40 turns. Not all that good i know
I will post it later.

Per


----------



## qqwref (Oct 8, 2010)

WTF are you talking about? Those are all turns relative to the inner cube. And the lowercase turns turn two adjacent layers at once.


----------



## mrCage (Oct 8, 2010)

qqwref said:


> WTF are you talking about? Those are all turns relative to the inner cube. And the lowercase turns turn two adjacent layers at once.


 
Ok, as I thought a notation confusion:tu I am used to upper case for the outer layers of the 5x5x5, and lower case for single inner slice turns of the 5x5x5. These slice turns are equivalent to turns of the outer layers of the inner 3x3x3. I wish there was a scripted Java applet available for this stuff. Maybe Randelshofer takes on the challenge??

Per

PS! Correctly understanding the notation your 2 sequences both cycle 3 centers of the outer 5x5x5 cube ...


----------



## qqwref (Oct 8, 2010)

You must still be misinterpreting, I've tried this sequence on CubixPlayer and it says it's SuperSolved. The corners of the 5x5 should never move.


----------



## mrCage (Oct 9, 2010)

Hmmm. That is weird. Well here is my notation. On the 5x5x5 we have layers R r M l L going from left to right. From top to bottom we have U u E d D. From front to back we have F f S b B. M turns same way as L. E same way as D and S same way as F. With this notation my complete and improved solution is:

1. l' u' (l m) d' (l'M') u' (l M) d (l' M') u2 l (11 turns block turn metric)

2. f l (f' S')d (f S) l (f' S') d' (f S) l2 f' (11 turns)

A full explanation will follow shortly. Part 1. and 2. are 2 block 3-cycles.

Per


----------



## qqwref (Oct 9, 2010)

I don't see why you consider it necessary to write 3x3 algs in 5x5 notation, but here you go:

u r' f2 r (S' b) r' f2 r (S b') u'
d' l b2 l' (S' f') l b2 l' (S f) d


----------



## mrCage (Oct 9, 2010)

qqwref said:


> I don't see why you consider it necessary to write 3x3 algs in 5x5 notation, ...


 
A matter of taste i guess. I always do the alg on a 5x5x5 to verify lack of side effect

Per


----------



## mrCage (Oct 9, 2010)

Sometimes easily solutions are buried in bad notation. I had the same 20 turn solution all the time. But i had written my block cycles with solely outer turns (on 3x3x3). Changing to proper commutator turns (with block turns in this case) was the way to go(to avoid side effects).

Coincidentally there is at least 18 different 20 turn solutions. 3 ways to cycle first 3 blocks. Then 3 ways for the last 3. In addition there are 2 ways to split it into 6 blocks ...

Per


----------



## qqwref (Oct 10, 2010)

Your solution is 22 moves - you used two setups, I used one. But yeah, a 20 move solution is probably the way to go.


----------



## mrCage (Oct 10, 2010)

What i meant is that my normal cube in cube is a 20 turn solution, involving 2 block 3-cycles. I realised i should "normalise" that one right after i posted my 22-turn solution. Oh well, you beat me to it cause of my "stupidity":tu

Check this thread!!

Per


----------

