Can I learn 5x5 full edge parity without algorithm?

[RubikZz]

Hallo,,

Can I learn the full edge parity of a 5x5 without any algorithm?
http://www.kungfoomanchu.com/guides/andy-klise-5x5x5-guide-v2.pdf
I saw anyone say that in a topic and I thought that should I know because I have the 5x5 now for a half year and I can do it only whit that tutorial.
And I will know the full edge parity because I'm than faster.
So has anyone a video or a tutorial, prefer a video where I can learn fast and without algorithms the 5x5?

Greetings Mathijs.

 

[Stefan]

No. You'll always use algorithms. Some are easier than others, though.

 

[Hershey]

You need only three algorithms.
One is r2 B2 U2 l U2 r' U2 r U2 F2 r F2 l' B2 r2, another is l' U2 l' U2 F2 l' F2 r U2 r' U2 l2, and lastly u2 r2 F2 u2 F2 r2 u2.
Everything else is completely intuitive.

(Post was edited a few times.)

 

[Stefan]

Why do you need two? There's just one parity.

 

[Hershey]

Why do you need two? There's just one parity.

Sorry, fixed. I learned 3 algorithms for edges.

 

[uberCuber]

Sorry, fixed. I learned 3 algorithms for edges.

You don't need all of them though.

EDIT: and that third case you listed isn't parity

 

[Hershey]

EDIT: and that third case you listed isn't parity

Oh.

 

[yockee]

One parity is when the two outer edges are switched on one edge, and the other parity is when two adjacent edges have one outer edge needing to switch with one from the other.

 

[Kirjava]

One parity is when the two outer edges are switched on one edge, and the other parity is when two adjacent edges have one outer edge needing to switch with one from the other.

They're just different cases of the same type of parity.

 

[yockee]

They're just different cases of the same type of parity.

I'm not sure if I explained it well, but the ones I was talking about are two totally different cases. One just involves one tredge, and the other involves two tredges. Two totally different algs. The first two that the guy a few posts up posted.

 

[Kirjava]

I'm not sure if I explained it well, but the ones I was talking about are two totally different cases. One just involves one tredge, and the other involves two tredges. Two totally different algs. The first two that the guy a few posts up posted.

Yep. They're the same type of parity.

 

[RubikZz]

If I do this algorithm the cube srambled complete:
u2 r2 F2 u2 F2 r2 u2, what do I wrong?

 

[Hershey]

If I do this algorithm the cube srambled complete:
u2 r2 F2 u2 F2 r2 u2, what do I wrong?

I think it was typed wrong.
The algorithm is
u2 r2 F2 (U2 u2) F2 r2 u2

 

[RubikZz]

Yeah thanks, these works.

 

[Christopher Mowla]

I think it was typed wrong.
The algorithm is
u2 r2 F2 (U2 u2) F2 r2 u2

Yes it was typed wrong, but it was not supposed to be that either. We're talking about the 5x5x5 here. Rather, u2 R2 F2 u2 F2 R2 u2. You can also perform it like this: (U2 u2) R2 F2 u2 F2 R2 (U2 u2).

In WCA notation in compact form, [Uw2 R2 F2: u2].

However, I prefer a related algorithm instead with a z cube rotation conjugate: [z Rw2 U2 F2: r2]

 

[Cheese11]

The first one is the one I use and it works fine for 5x5 and 4x4 edge paritys. Plus, it's easy to memorize