Simple Little Idea for Higher-order Cube Edge Pairing

[CuBeOrDiE]

I've just been thinking... when you solve (or at least I solve) bigger cubes, I use reduction. But when you are pairing up edges, you temporarily break up centers, and occasionally, i find i might mess them up. So instead, why not solve two opposite centers, put them on the U and D side, and then solve 6 edge pairs, placing them near those centers. That way, you don't worry about messing up the centers in the middle. Then, solve the remaining 6 centers and edges.

This has probably been thought of, but I found no thread about it. It's a simple concept, and won't improve times a lot, but I find it helps. What do you think?

 

[MichaelP.]

Would that mean that you would be able to do R, R', L, L', F, F', B, B' , because that would place the slice where you are solving your centers?

 

[Robert-Y]

Erm... wouldn't it be very difficult (move consuming) to preserve those 8 tredges whilst solving the rest of the centres?

 

[CuBeOrDiE]

Would that mean that you would be able to do R, R', L, L', F, F', B, B' , because that would place the slice where you are solving your centers?

yes, that was the point of not solving centers. It would actually SAVE moves on that (just a bit) and then keep the move amount for centers the same. do you think im right?

 

[Kirjava]

Poorly thought out idea is poorly thought out.

lern2yau.

 

[MichaelP.]

When solving the last 4 centers, you'll probably have to mess up the tredges, because you can't keep them near the centers and solve the other four centers.

 

[CuBeOrDiE]

your comment is poorly thought out because it doesn't say WHY my proposition is poorly thought out.

ironic?

 

[Robert-Y]

Poorly thought out idea is poorly thought out.

lern2yau.

Ahem... lern2K5 you mean...

 

[CuBeOrDiE]

When solving the last 4 centers, you'll probably have to mess up the tredges, because you can't keep them near the centers and solve the other four centers.

erm... we only solve 8 tredges. not all 12.

CLARIFICATION:

1-solve two opposite centers and put them in the U and D
2-solve 6 tredges. this is easy because you don't worry about breaking up the middle four centers. make sure to put them in the U and D slice
3-solve remaining four centers
4-solve remaining 6 tredges
5-solve like 3x3

the advantage is those 8 tredges take a few less moves to solve because you have no centers to mess up. and even a few moves make a difference...

hope that clears it up

 

[Sa967St]

I've just been thinking... when you solve (or at least I solve) bigger cubes, I use reduction. But when you are pairing up edges, you temporarily break up centers, and occasionally, i find i might mess them up. So instead, why not solve two opposite centers, put them on the U and D side, and then solve 8 edge pairs, placing them near those centers. That way, you don't worry about messing up the centers in the middle. Then, solve the remaining 4 centers and edges.

This has probably been thought of, but I found no thread about it. It's a simple concept, and won't improve times a lot, but I find it helps. What do you think?

Have you even tried doing a solve like that yet?

 

[Robert-Y]

Here's what I thought you meant:

1. Solve 2 opposite centres
2. Solve 8 tredges and place them next to the two opposite centres.
3. Solve the remaining 4 centres
4. Solve the remaining 4 tredges

Correct?

 

[aronpm]

Let me make this extremely clear.

If you have the 8 'tredges' solved on the U and D faces, you can only make E slice turns. You cannot turn the FBRL faces. You cannot solve the centres.

 

[DT546]

Let me make this extremely clear.

If you have the 8 'tredges' solved on the U and D faces, you can only make E slice turns. You cannot turn the FBRL faces. You cannot solve the centres.

you can, it just won't be very efficient

 

[Kirjava]

your comment is poorly thought out because it doesn't say WHY my proposition is poorly thought out.

I guess you're too stupid to realise why this is a bad idea.

lern2yau.

Ahem... lern2K5 you mean...

With edge pairing? XD

EDIT; the intention is the same, solve three (or six if you're crazy) tredges and the centres arn't balls.

 

[shelley]

Let me make this extremely clear.

If you have the 8 'tredges' solved on the U and D faces, you can only make E slice turns. You cannot turn the FBRL faces. You cannot solve the centres.

Well, you can do double turns, but that still limits your options a lot.

And that is why poorly thought out idea is poorly thought out.

 

[CuBeOrDiE]

oh, i was always saying 8 tredges?! i meant 6!!! where was my head? i c what ur saying. my bad for doing...that...

:fp forgive me i had one of those moments. let me go fix my posts

 

[Kirjava]

oh, i was always saying 8 tredges?! i meant 6!!!

Sure you did ^_^

 

[CubesOfTheWorld]

I just want to know how to pair up the last 4 centers without screwing up the edges.

 

[puzzlemaster]

oh, i was always saying 8 tredges?! i meant 6!!! where was my head? i c what ur saying. my bad for doing...that...

:fp forgive me i had one of those moments. let me go fix my posts

I don't think that that really fixes the problem with your method..

 

[CuBeOrDiE]

oh, i was always saying 8 tredges?! i meant 6!!!

Sure you did ^_^

no, i did a few solves with solving 6. but for some reason i was thinking 8 when i posted this. my bad for looking like a fool.

but i can solve 5x5 centers in like 85 moves, so im not completely retarded