# What's the difference between 4x4 and 5x5?



## UnicornPow3r (Jul 18, 2011)

Yeah so, I'm not sure if this is the correct question, but searching the web I haven't really found any straight answers to this, I have only become more confused.

As of right now I sit on a Eastsheen 4x4 (which I think locks up waaaay to much) and a QJ 5x5 (which is heavy, and turns horribly in my opinion) which I decided to learn to do after just having them for a couple of months.

I just finished learning the 4x4 I think, but I won't start on the 5x5 just yet, because I need to drill the new algorithms ( not to many, and not to hard, I am happy about that), so I don't want do dive into the 5x5 just yet, therefore I figured I'd just ask what the appearent differences are.

Of course, I understand that the 5 goes back to fixed centers, which greatly pleases me, but except for that (making the centers that is), is there any new algorithms, or do I know how to solve all/most of the 5x5 by knowing the 3x3 and 4x4 from beginning to end?

I solve the 3x3 by doing cross, F2L, 2-Look OLL and 2-Look PLL, and use mostly the same for the 4x4, + the extra edge parity algorithm. 

A bit long post, I know, but any things I should know about the 5x5 (and beyond! Already ordered a V6 and V7) would be greatly appreciated! 

Cheers from Norway!
Kristoffer


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## emolover (Jul 18, 2011)

There is not much difference between 4x4 and 5x5 solving for the most part. The only real difference is the parities. 

You should not be using algorithms to solve big cubes unless it is parity. I have never used algorithms.


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## uberCuber (Jul 18, 2011)

If you can solve a 4x4, you don't need any more algs for 5x5 and up. The centers can be solved completely intuitively, though it will probably take you awhile to get used to. The edges are intuitive in the same way they are on 4x4. Parity can occur during edge pairing, but can be fixed using a 4x4 OLL parity alg.


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## Mike Crozack (Jul 18, 2011)

on the 5x5, theres one new algorithm, and one algorithm from 4x4 thats applied differently. 
when pairing edges, you may end up with 2 unsolved ones, to move on edge-wing piece to the other edge, put it on your right and do l' U2 l' U2 F2 l' F2 r U2 r' U2 l2
and the other algorithm is to flip two wings in the same edge, which is the oll parity alogrithm from 4x4


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## cityzach (Jul 18, 2011)

one has 4 layers and one has 5 layers xD


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## UnicornPow3r (Jul 18, 2011)

emolover: I am not sure what you mean by no algorithms should be used. Do you mean the whole cube is intuitive? I dunno if I solve it in a less optimal way, but doing it like I solve the 3x3 does use the same algorithms. If you would elaborate, it would be most satisfactory :]

uberCuber: Ah, this was actually what I wanted to hear, I was a bit afraid that I needed to learn way more complicated algorithms than I use for the 4x4. Say for example, when I solve the edges in the 4x4, and I am down to the last pair I put the same colors next to each other and use the [Dd, R, F', U, R', F, Dd'] algorithm to solve it, is this the same for 5x5? Seeing as there is a middle edge piece as well I would think there would be something else needed, but as I haven't tried I wouldn't know.


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## Clayy9 (Jul 18, 2011)

You don't need to know any new algorithms for the 5x5, but knowing a parity alg (for 1-edge parity) can be useful, but not necessary. I've solved the 5x5 using no algorithms other then for the 3x3 part. 

The parity alg is a bit different then for the 4x4, but you CAN solve the cube without using it.


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## Kirjava (Jul 18, 2011)

hm.

midges on 5x5x5 ELL is interesting if you don't solve them seperately. that's the only real difference in direct solving methods though.

the perceived most efficient piece solving order seems to change at about N>6 for K4.


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## AustinReed (Jul 18, 2011)

Mike Crozack said:


> on the 5x5, theres one new algorithm, and one algorithm from 4x4 thats applied differently.
> when pairing edges, you may end up with 2 unsolved ones, to move on edge-wing piece to the other edge, put it on your right and do l' U2 l' U2 F2 l' F2 r U2 r' U2 l2
> and the other algorithm is to flip two wings in the same edge, which is the oll parity alogrithm from 4x4


 
I don't even use that algorithm, let alone any new ones.


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## majikat (Jul 19, 2011)

> What's the difference between 4x4 and 5x5?


9.
hahaha!

Seriously, though. Like everyone said, once you've learned to solve a 4x4, you pretty much have the necessary knowledge to solve any larger cube...the most basic method is always the same- solve centers, solve edges, solve cube using a 3x3 method.


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## uberCuber (Jul 19, 2011)

NOTE: All comments on this post are assuming standard reduction method, so don't bother responding with direct solving method info.




Mike Crozack said:


> on the 5x5, theres one new algorithm


 
Or not. I did not learn one single new algorithm when I first got my 5x5, and solved it quite successfully. It was only very recently (i.e. after I was already sub-2) that I started learning any other algs. The case you mention can be solved with a combination of a [d R F' U R' F d'] type of thing (though I do it on M slice, not E) and the 4x4 OLL parity alg.



Clayy9 said:


> The parity alg is a bit different then for the 4x4



That depends on what alg you use for 4x4. I use _exactly_ the same alg for both.



UnicornPow3r said:


> Do you mean the whole cube is intuitive?



The centers should be solved 100% intuitively. The first 10 out of 12 edge groups are always done intuitively, and even the last two edges are sometimes intuitive as well; however it can be useful to learn some algs for the last two edges for those cases that aren't super intuitive.



> Say for example, when I solve the edges in the 4x4, and I am down to the last pair I put the same colors next to each other and use the [Dd, R, F', U, R', F, Dd'] algorithm to solve it


 
What you need here is to understand what that algorithm is doing intuitively. Notice that the R F' U R' F part flips the edge pair in the FR position. I think you can take it from there.
As long as you understand what your alg is doing, you can always solve the last two edges on 5x5 with no more than two applications of the alg you mentioned (and/or a mirror of some kind), and no more than one application of the 4x4 OLL parity alg.


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## penfold1992 (Jul 19, 2011)

i have an issue understanding certain comms and found the centers difficult on a 7x7 at first. the first 7x7 attempt was with Yau's brand new 7x7 at the uk open however i had developed a muscle injury in my right hand and i couldnt turn the cube well at all so i got frustrated with losing what i was doing.

if you can solve a 4x4 there is nothing new to learn as such, maybe just more understanding of how a rubiks cube works is needed. if your using a 4x4 method that works for you yet arent sure what it actually does i recommend working out for yourself how certain algs do certain things etc... other then that 5x5 is an awesome puzzle, not too fast to make it "easy" but not to slow to solve to make it tedious xD


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## tx789 (Jul 19, 2011)

The reduction method translates up to infinity by infinity


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## uberCuber (Jul 19, 2011)

penfold1992 said:


> i have an issue understanding certain comms and found the centers difficult on a 7x7 at first.


 
I feel like I say this kind of thing too often, but centers should be intuitive to the point that you don't even need to understand or use comms at all. I never (emphasis: never) use comms during 7x7 centers. Not being good at comms is not a good excuse for not being good at (sighted) 7x7 solves.


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## UnicornPow3r (Jul 19, 2011)

ubercuber: Wow! Thank you for this much informative post! I actually didn't know what the algorithm actually did, so it is possible to achieve different results on a 5x5 by, for instance, changing the Dd and Dd' to just d and d'? Like I said, I haven't started the 5x5 and I don't have it with me at the moment, so I can't try to see what it would accomplish. A lot more responses to this thread than I expected though, and I must say I am excited to start the 5x5 now, after being a bit reluctant because I thought it would be so much work.


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## TMOY (Jul 20, 2011)

uberCuber said:


> I never (emphasis: never) use comms during 7x7 centers.



I'm 100% certain (emphasis: 100% certain) that you do.

Hint: Rw U Rw' U' and similar sequences of moves are comms.


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## Kirjava (Jul 20, 2011)

I wanna see him do L2Centres without comms.

Maybe it's possible idk. I'll try when I get home from work.


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## uberCuber (Jul 20, 2011)

Crap wasn't thinking about little things like sexy move. I was thinking of such things as 3-cycles and stuffs like that...

Kir I can do an example with alg.garron.us and post the link here later. It'll be awhile though because ima go running rn.


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## Kirjava (Jul 20, 2011)

Cool. I'm looking forward to seeing how you solve a single oblique without using niklas.


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## uberCuber (Jul 20, 2011)

L2C Scramble: 3r' F l F2 r F' l' F' 3r U2 l' U' l 3r' F 3r r' F 3l' U2 3r U' 3r' r U2 r' 3l2 F' l' 3l' U l

3l F' 3l'
3l' U2 3l r U 3l' U2 3l U2 r' U l' U2 l
r U 3r r' U 3r' r U' r' 3l l' F2 3l' l
F' U' 4l' U l U' 4l F2 l' U2 l' U l U' r' F2 r



L2C Scramble: l F' l' 3l' U' 3l F' 3r U' r' U2 3r' F r 3l F' 3l' l' U 3l F' l 3l' F' 3r U 3r' F' r U2 r' 3l F' 3l'

U' 3r' F r F r' F' 3r2 U2 3r'
U' 3l' U 3l U r U2 r'
U' 3l' U 3l U l' U' l U' 3r' F2 3r
l' U' l F 3r U r' U' 3r' r U2 r U' r' U r U2 r'


There's probably some stupidity present in this that I'm too tired to notice. Feel free to point it out if you find it.


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## Kirjava (Jul 20, 2011)

There are some comms in the examples you posted.

Anyway, what I want to know is how you solve this case;







EDIT: Ok, I found a way of doing this without comms. Comms are more efficient though.


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## uberCuber (Jul 20, 2011)

Kirjava said:


> There are some comms in the examples you posted.



Wait, are you referring to the 4-movers, or am I actually doing longer comms without even realizing it?



> Anyway, what I want to know is how you solve this case;
> 
> 
> 
> ...


 
With the way I solve, that case doesn't happen to me, but if it did, I would take around 10 moves to solve it (depending on what turn metric is being used) - (F) 3r U' r' U (3r' r) U' r U r' . I would never claim my way of solving the last row of L2C is more efficient that anything, I just solve according to my turning style.

EDIT: holy **** I just realized that's a comm, [3r r', r U' r' U]. I'm going to facepalm myself now. :fp :fp :fp. In fact, that's such an incredible fail I'm going to add it to my sig.
Well I guess that reinforces the idea of how intuitive comms are...


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