- Dec 23, 2009
- #1
R
rpgragexp♀
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- Dec 21, 2009
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- 5
I wanna know which one is easier to solve. Some people told me that 4x4x4 is easier because it has less cubes, but then others tell me the 5x5x5 is easier because it has a center piece or something. Can anyone tell me which one is easiest to solve? I already can solve the 3x3x3 so I want something more challenging but not too challenging(Kinda like the next level, after 3x3x3)
- Dec 23, 2009
- #2
D
daniel0731ex
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- Dec 10, 2008
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5x5.
- Dec 23, 2009
- #3
Z
Zarxrax
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4x4x4 isn't all that challenging except for the parity cases, which are pretty darned hard to work out on your own unless who have a knack for making your own algorithms intuitively.
5x5x5 should be easier, since you wont run into those parity cases.
- Dec 23, 2009
- #4
IamWEB
Member
5x5x5 is easier, because it has defined centers so that you won't mess that up, and you don't get those parity cases.
4x4x4 center-making and edge building should be easier though.
So I'd just say go with 4x4x4, because it really is the 'next level up.'
- Dec 23, 2009
- #5
~Phoenix Death~
Member
4x4x4- No centers, parity,
5x5x5-Has a center, HAS parity (though different), need to pair up more edges, need to build up centers that are larger, and yeah
- Dec 23, 2009
- #6
gibsonguitarist55
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- Jul 16, 2009
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- 59
5x5 and also my favorite
- Dec 23, 2009
- #7
Cyrus C.
Member
What's the 5x5 parity?
- Dec 23, 2009
- #8
michaellahti
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- Sep 24, 2009
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I'd say 5x5 is easier, although I'd get a 4x4 first if I were you.
- Dec 23, 2009
- #9
EmersonHerrmann
Member
Cyrus C. said:
What's the 5x5 parity?
The mid-middle cubie/wing edges are flipped. I think.
Edit: 4x4 all the way fo sho xD
- Dec 23, 2009
- #10
Cyrus C.
Member
That's not considered parity is it?
- Dec 23, 2009
- #11
masterofthebass
Premium Member
5x5 parity is the same as 4x4 parity. You can just see it earlier...
- Dec 23, 2009
- #12
~Phoenix Death~
Member
It's not necessarily a Parity. It's what Rob said. Rob H206 I think.
The first picture shows a prity that can be fixed using the 4x4 Parity.
The second picture shows a picture that can be fixed with
[(Ll)' U2 (Ll)' U2] F2 [(Ll)' F2 (Rr)] [U2 (Rr)' U2 (Ll)2]
- Dec 23, 2009
- #13
capoboy
Member
either one is fine. 4x4 is good, 5x5 is also good. Basically, they have the same principle, solve the center, the edges, and solve like 3x3. Maybe you can get megaminx if you want to, that would be more challenging because it has different shape. lol.
- Dec 23, 2009
- #14
Cyrus C.
Member
Okay, I didn't know it was considered parity since you can see it before the 3x3 step.
Oh & sorry for hi-jacking this thread.
- Dec 23, 2009
- #15
~Phoenix Death~
Member
Try applying K4 to both. I find K4 is actually fun. Fun as is whopee.
- Dec 23, 2009
- #16
PEZenfuego
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- Jul 21, 2009
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One of my friends told me the basics of how to solve a 5x5 with reduction method. He told me to build the centers, build the tredges (he didn't use that word) and then solve like a 3x3. With that knowledge, it didn't take me very long to figure out how to solve it. With that said, I wouldn't have had a popsicle chance in hell figuring it out on a 4x4 due to parity. So figuring out how to solve a 4x4 is harder.
I would consider 4x4 easier to solve for a few reasons. I can solve a 4x4 much faster than a 5x5 since I know the parity cases, the color scheme, and it has less cubies than a 5x5. In terms of solving the cube, it is easier.
Of course it is all a matter of opinion...
- Dec 23, 2009
- #17
LNZ
Premium Member
For me 5x5 is much easier. Despite the longer solving time. There is no top layer parity issues on a 5x5.
There is parity ona 5x5 though. You get it in edge pairing. But you use one alg from the 4x4 and only 1 new one for the 5x5 to get past it.
- Dec 23, 2009
- #18
E
ElderKingpin
Member
Off-topic: I think a parity is anything that wouldnt make any sense on a normal 3x3
On-topic: I think 5x5 is easier, because it doesnt have as many parities as a 4x4 (but if you plan to go on to the 6x6 get a 4x4 first) the even number cubes have no fixed centers and more parities then the odd ones.
- Dec 23, 2009
- #19
PEZenfuego
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- Jul 21, 2009
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LNZ said:
For me 5x5 is much easier. Despite the longer solving time. There is no top layer parity issues on a 5x5.
There is parity ona 5x5 though. You get it in edge pairing. But you use one alg from the 4x4 and only 1 new one for the 5x5 to get past it.
You see the parity on a 5x5 prior to finishing F2L (er uh...F4L) and that makes it a much smaller pain in the rear (in my opinion). I used to just do (Rr U2)x4 when I hit two opposite tredge parity. Granted, this didn't fix the parity, but it eliminated the possibility of getting it again...
- Dec 23, 2009
- #20
~Phoenix Death~
Member
LNZ said:
For me 5x5 is much easier. Despite the longer solving time. There is no top layer parity issues on a 5x5.
There is parity ona 5x5 though. You get it in edge pairing. But you use one alg from the 4x4 and only 1 new one for the 5x5 to get past it.
Look at the top of the page, there's a second.
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