Is a 6x6 cube algorithmically redundant?

[mband]

Hi everyone, I'm considering replacing my old Rubik's-brand 3x3 and 5x5 with a new set of nicer cubes, including 2x2, 3x3, 4x4, and 5x5. My question is: should I get a 6x6 as well (and/or a 7x7), or is a 6x6 redundant if I get the smaller ones?

My primary interest is in figuring out how to solve these cubes myself, but if solving the 6x6 is simply a straightforward repetition of the algorithms for the 4x4 and/or 5x5, then I'd consider the 6x6 redundant and would have little interest in doing it (aside from the pride factor). On the other hand, if there are some unique things that need to be figured out for the 6x6 that aren't straightforward extensions of the smaller cube sizes, then I'd want to get one.

Any opinions on whether there's unique algorithms that need to be figured out for the 6x6, or is the 6x6 redundant from an algorithmic point of view?

Note, I don't want to just "look up" the solutions, as that would spoil the fun for me in figuring it out for myself.

Thanks!

 

[benskoning]

the 6x6 is different because you have to pair inner edges which is a bit more of a chalange. 7x7+ is really your choice.



Welcome to the forum.

 

[panyan]

a 6x6 is basically a large version of a 4x4 imo, its rather boring in my opinion

I really like my 7x7, even though it is basically a large 3x3 becuase it takes longer to solve so is more satisfying

I just got an 11x11 and i feel great when i solve it, even if you do reduce it to a 3x3 in solving

 

[qqwref]

Solving the 6x6x6 is sort of similar to the 4x4x4, but the centers are a lot more complicated. The existence of oblique centers definitely adds something new.

Also, you can simulate tons of cuboids on the 6x6x6, but not on the 7x7x7 That is one cool advantage of even cubes.

 

[mband]

Thanks for the quick replies, everyone, and for welcoming me to the forum. The fact that the 6x6 can simulate all the smaller cubes is an interesting observation. It sounds like there's at least a little extra figuring out for the 6x6, so I'll likely get one. From the reviews I've read, the Shengshou V3 or V4 is the one I should look at getting. Anyone have any experience ordering from here:
http://www.aliexpress.com/item/magi...layer-6x6x6-magic-puzzle-white/726297604.html

 

[MWilson]

Thanks for the quick replies, everyone, and for welcoming me to the forum. The fact that the 6x6 can simulate all the smaller cubes is an interesting observation. It sounds like there's at least a little extra figuring out for the 6x6, so I'll likely get one. From the reviews I've read, the Shengshou V3 or V4 is the one I should look at getting. Anyone have any experience ordering from here:
http://www.aliexpress.com/item/magi...layer-6x6x6-magic-puzzle-white/726297604.html

Am I nuts or does that 6x6 look like eight 3x3s stacked together.

 

[Christopher Mowla]

If you haven't solved a 6x6x6 yet, then it might be good to solve a virtual cube first because you can test algorithms that you come up with on a solved cube and then use them when you are solving a scrambled cube. Gabbasoft is a good one. I solved my first 6x6x6 without help by solving some online java applet (I didn't actually buy a 6x6x6 for a few months after I already solved a virtual one). So if you want to buy new/better smaller cubes, you can and still you can learn the 6x6x6.

 

[benskoning]

Thanks for the quick replies, everyone, and for welcoming me to the forum. The fact that the 6x6 can simulate all the smaller cubes is an interesting observation. It sounds like there's at least a little extra figuring out for the 6x6, so I'll likely get one. From the reviews I've read, the Shengshou V3 or V4 is the one I should look at getting. Anyone have any experience ordering from here:
http://www.aliexpress.com/item/magi...layer-6x6x6-magic-puzzle-white/726297604.html

I would not recommend that store. try some place like hknowstore.

 

[Kirjava]

Also, you can simulate tons of cuboids on the 6x6x6, but not on the 7x7x7

You can always ignore layers.

 

[qqwref]

Ignoring the middle layer is kind of awful. Have you tried it? This one time I did a 4x4x4 solve on a 5x5x5 and it took me like an extra 30 seconds (and felt weird to boot).

 

[mband]

You can always ignore layers.

The way I've figured it, a 6x6x6 can simulate a 5x5x5 (if you don't do middle cuts), 4x4x4 (don't do 2nd or 4th cuts), 3x3x3, as well as 2x2x2. But an odd-ordered cube can only simulate smaller odd orders, so, e.g. a 7x7x7 couldn't simulate a 4x4x4, unless I'm missing something.

Thanks for the other advice too - I'll give gabbasoft a whirl.

As for hknowstore, I don't see Shengshou 6x6x6's on it, unfortunately - otherwise it looks like a great source for cubes.

 

[mark49152]

Might be a n00b question, but why doesn't the ignored layer affect the other layers? E.g. parity?

 

[mband]

This one time I did a 4x4x4 solve on a 5x5x5 and it took me like an extra 30 seconds (and felt weird to boot).

How did you simulate a 4x4x4 on a 5x5x5, if I may ask? I didn't think that was possible (see my previous post).

EDIT: I think I understand what you mean now, by ignoring the center layer while solving. I don't think I'd like that at all.

 

[AvGalen]

A 6x6x6 IS algorithmically redundant if you look at the actual algorithms. But it is actually the part that you solve without algorithms (the centers) that are still a challenge.

I often do 2345678 relays and every cube is different. So different actually that if I solve 2 8x8x8's in a row I get good solves, but if I solve a 7x7x7 directly after that I will "forget' that it has fixed centers and get a bad time

 

[Christopher Mowla]

Might be a n00b question, but why doesn't the ignored layer affect the other layers? E.g. parity?

Well for common parity algorithms, just as you cannot affect a portion of the 3x3x3 cube using only R and U moves, the move optimal commutators found within common parity algorithms ("single slice turn based algorithms") only need to run through certain layers to get the job done.

There are parity algorithms that exist which need to use all layers to affect wing edges in only one layer, however.

For example, here's the "Holy Grail" one edge flip alg.'s unabridged commutator and conjugate outline:
3r' e R 3l' 3u 3U'
3U 3u' 3l 3u' 3f' 3U' 3f 3u 3l' 3u
3u 2U' 3f' 3U' 3f 2U 3u'
3U 3u' 3l R' e' 3r

In SiGN notation, 3U turns the inner-inner layer, and 2U turns the outer-inner layer. Since this algorithm affects the inner orbit (the wing edge pieces in the inner-inner layer of the 7x7x7 are affected by the algorithm..."where 3U reaches"), if we omit the 2U' and 2U moves "which affect the outer-inner layer",

3r' e R 3l' 3u 3U'
3U 3u' 3l 3u' 3f' 3U' 3f 3u 3l' 3u
3u 3f' 3U' 3f 3u'
3U 3u' 3l R' e' 3r

the oblique centers in the inner-inner layer are affected. (But one can also view these centers as being in the outer-inner layer as well...but you still can view the centers affected as the inner-inner layer/the other layer).

So even with this extreme example that might seem to "violate" the rule you stated, it technically doesn't, but it can be viewed as doing just that.

Summary
An algorithm can only affect layers that it runs through, but since oblique centers can appear to be in either the inner-inner layer or outer-inner layer, some may consider algorithms which use a slice turn which is different that the slices containing wing edges that are affected to affect the slices in which the wing edges are affected.

 

[Kirjava]

How did you simulate a 4x4x4 on a 5x5x5, if I may ask?

By ignoring layers?

 

[CarlBrannen]

In addition to different algorithms for different sizes, it's inevitable that different cubes feel differently. So your best finger tricks are likely to be different for different size cubes, even for the same algorithm.

And you can "ignore layers" on a 5x5 to get a 4x4x4, but you should also peel off the stickers as otherwise it's telling you which side is which.

 

[mark49152]

~snip~

Thanks. I think you're a few steps ahead of me though. I'm new to big cubes and was simply thinking about solvable permutations. Is it ever the case that you can't reach a "solved" state with the outer four layers because the permutation of the middle layers prevents it? For example, even number of edges flipped?

 

[peterbone]

I worked out how to solve the 3x3x3 and large cubes myself using a layers method. The 6x6x6 and above was not much different after the 5x5x5.

 

[mband]

Is it ever the case that you can't reach a "solved" state with the outer four layers because the permutation of the middle layers prevents it? For example, even number of edges flipped?

You should always be able to solve the outer four layers. This is simply a subset of all 5 layers, which is also solvable. CarlBrannen's comment re peeling off the stickers is also valid - if you don't, then the center stickers give you valuable orientation information. It may suffice to simply peel off the center stickers rather than the stickers from all the middle layers - someone with more experience can probably verify exactly which stickers need to be peeled off.