# Simple Little Idea for Higher-order Cube Edge Pairing



## CuBeOrDiE (Apr 13, 2010)

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?


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## MichaelP. (Apr 13, 2010)

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?


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## Robert-Y (Apr 13, 2010)

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


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## CuBeOrDiE (Apr 13, 2010)

MichaelP. said:


> 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?


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## Kirjava (Apr 13, 2010)

Poorly thought out idea is poorly thought out.

lern2yau.


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## MichaelP. (Apr 13, 2010)

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.


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## CuBeOrDiE (Apr 13, 2010)

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

ironic?


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## Robert-Y (Apr 13, 2010)

Kirjava said:


> Poorly thought out idea is poorly thought out.
> 
> lern2yau.



Ahem... lern2K5 you mean...


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## CuBeOrDiE (Apr 13, 2010)

MichaelP. said:


> 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


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## Sa967St (Apr 14, 2010)

CuBeOrDiE said:


> 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?


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## Robert-Y (Apr 14, 2010)

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?


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## aronpm (Apr 14, 2010)

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.


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## DT546 (Apr 14, 2010)

aronpm said:


> 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


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## Kirjava (Apr 14, 2010)

CuBeOrDiE said:


> 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.



Robert-Y said:


> Kirjava said:
> 
> 
> > lern2yau.
> ...




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.


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## shelley (Apr 14, 2010)

aronpm said:


> 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.


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## CuBeOrDiE (Apr 14, 2010)

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


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## Kirjava (Apr 14, 2010)

CuBeOrDiE said:


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




Sure you did ^_^


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## CubesOfTheWorld (Apr 14, 2010)

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


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## puzzlemaster (Apr 14, 2010)

CuBeOrDiE said:


> 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..


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## CuBeOrDiE (Apr 14, 2010)

Kirjava said:


> CuBeOrDiE said:
> 
> 
> > oh, i was always saying 8 tredges?! i meant 6!!!
> ...



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


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## CuBeOrDiE (Apr 14, 2010)

puzzlemaster said:


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



put 3 on top and 3 on bottom. then line them up to where you have a face you can easily turn. dunno, did i explain that well enough?

edit: this is not a method, just an idea. ive only done a few solves like this


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## MichaelP. (Apr 14, 2010)

puzzlemaster said:


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



I think if you had one unpaired edge in the top and bottom, you could do FBRL turns by positioning the unpaired edges correctly. It won't be efficient though.


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## CuBeOrDiE (Apr 14, 2010)

MichaelP. said:


> puzzlemaster said:
> 
> 
> > CuBeOrDiE said:
> ...



its just an idea. but if you could solve edges and the 3x3 in one go, that would be awesome. but i haven't found a way yet, prbly because there isn't one.

edit: FYI my 5x5 has a snapped core so i can't solve right now  but im getting vcubes!


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## Kirjava (Apr 14, 2010)

CuBeOrDiE said:


> 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.




You *edited your posts to make it seem like this was your idea all along*. Douchebag move. There was no mention of doing only 6 until I brought it up. You kept saying there were four edges to complete after the centres were complete. You made this mistake again and again and again. You would've noticed that you'd put 8 and 4 instead of 6 and 6 if you actually meant that. 

You did it SO MANY TIMES. You did it in your CLARIFICATIONS.

I do not believe you.


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## koreancuber (Apr 14, 2010)

Bad idea, but stop scorching him. He just wasn't thinking enough.


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## CuBeOrDiE (Apr 14, 2010)

Kirjava said:


> CuBeOrDiE said:
> 
> 
> > 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.
> ...



w/e. i don't have a 5x5 in my hands right now, so i didn't think when writing this. that is the naked truth. but when i solved like this before my 5x5 was broken i DID solve 6 tredges.

edit: i did that so new people to this thread wouldn't post the same things you posted. just to save time

editing the edit: i changed my posts so others wouldn't waste our time saying things like "hey, that doesn't work!"


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## Robert-Y (Apr 14, 2010)

I thought of this idea before. After the 6 tredges and all of the centres have been solved, I was thinking about doing F M or F' M, then solving the 6 remaining tredges by using 2 wing chain solving. (Use the last two tredges algs if you need to...)

(Rotate the cube so that the 6 tredges which have already been solved are on the LF, LD, LB, RF, RD, and RB slots before you do the F M or F' M.)

I still can't seem to find a solution for the high move consumption though :/ ...


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## waffle=ijm (Apr 14, 2010)

lol stadler for 5x5 without actually making blocks.


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## cuBerBruce (Apr 14, 2010)

I don't necessarily completely solve the last 4 centers (before starting edges), but have at least horizontal lines on each of those centers, and preserve horizontal lines on those centers while solving the first 8 edge groups. Then I solve the last 4 centers before doing the last 4 edge groups.


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## CuBeOrDiE (Apr 14, 2010)

Robert-Y said:


> I thought of this idea before. After the 6 tredges and all of the centres have been solved, I was thinking about doing F M or F' M, then solving the 6 remaining tredges by using 2 wing chain solving. (Use the last two tredges algs if you need to...)
> 
> (Rotate the cube so that the 6 tredges which have already been solved are on the LF, LD, LB, RF, RD, and RB slots before you do the F M or F' M.)
> 
> I still can't seem to find a soution for the high move consumption though :/ ...


 
i will w/e i get a my 5x5  wish me luck


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## Robert-Y (Apr 14, 2010)

cuBerBruce said:


> I don't necessarily completely solve the last 4 centers (before starting edges), but have at least horizontal lines on each of those centers, and preserve horizontal lines on those centers while solving the first 8 edge groups. Then I solve the last 4 centers before doing the last 4 edge groups.



I think Oliver Perge does the same thing btw.

I wonder how many moves are saved on average for solving the first 8 tredges like this and how many moves are required to restore the 4 centres...


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## qqwref (Apr 14, 2010)

You can do this with pairing 6 edges. The problem is that only being allowed to use r, 3r, ..., 3l, l, U moves severely limits how efficient your centers can be, and since on larger cubes centers are most of the solve this will slow you down in the long run.

A related idea I had a while ago was to make all centers, and then allow basically yourself to mess them up in strips while solving the first 8 edges. So the idea there is you can do R2, F2, B2, L2 at any time no matter what the centers look like, and this should save a lot of moves on those edges. Then you can either fix the centers (only 4(N-2) pieces, it's quick) and do the edges, or - if you've left only the middle four edges to be solved - you can actually solve the middle part like a 3x3x(N-2) cuboid (well, if N is even, otherwise you have some flipped edges in the middle layer), which is probably faster since you can blockbuild and use domino algs and stuff. I haven't tried this enough to know if it can be as fast as (or faster than?) normal redux on big cubes, but it doesn't seem obviously slower.


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## Dene (Apr 14, 2010)

I thought of this quite a while ago.

Of course, at the time I didn't think of doing only the first 6 edges instead of 8 (silly me). 
As you can see from the responses then, this has been thought of many times before in the past and has always been rejected.

In fact, PatrickJameson came out with the same idea a bit later.

My response to this method: I would not use it for anything bigger than 4x4. It could potentially be fast, but I don't think it can be faster than standard reduction.


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## CuBeOrDiE (Apr 14, 2010)

Dene said:


> I thought of this quite a while ago.
> 
> Of course, at the time I didn't think of doing only the first 6 edges instead of 8 (silly me).
> As you can see from the responses then, this has been thought of many times before in the past and has always been rejected.
> ...



yeah, im starting to think the same thing


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## Kirjava (Apr 14, 2010)

koreancuber said:


> Bad idea, but stop scorching him. He just wasn't thinking enough.




I'm not scorching him for his idea.



CuBeOrDiE said:


> edit: i did that so new people to this thread wouldn't post the same things you posted. just to save time




They don't have the context that we have, this thread would get confusing otherwise. Luckily, I have provided it.

I still don't believe you, but that doesn't change anything - method discussion is fun regardless. ^_^



Robert-Y said:


> I still can't seem to find a solution for the high move consumption though :/ ...




The 5x5x5 is a tricky beast. I've thrown all sort of weird ideas at it and nothing compares to redux. 

The closest thing I found (aside from k4) is something like this;

LR centres
1x4x5 block or three 'cross' tredges
M centres
complete 4x4x5 block with intuition
complete F4L with intuition/pairing
pure LL

Then there's the n>4 k4 variation where you do all centres first, pair the cross, F2L with corners and midges and direct solve the rest. I'm not a fan.

And this is why my 5x5x5 times suck.


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## Robert-Y (Apr 14, 2010)

One day, some someone will find and develop a method which is greater than reduction... maybe...


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## CuBeOrDiE (Apr 14, 2010)

Kirjava said:


> koreancuber said:
> 
> 
> > Bad idea, but stop scorching him. He just wasn't thinking enough.
> ...



What are your 5x5 times?


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## Kirjava (Apr 14, 2010)

~2:00


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## Dene (Apr 14, 2010)

Perhaps clicking the WCA profile button would help?


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## Kirjava (Apr 14, 2010)

You'd think that, but my results from the last competition are wrong for 5x5x5 XD


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## miniGOINGS (Apr 14, 2010)

waffle=ijm said:


> lol stadler for 5x5 without actually making blocks.



Mhm, like a bad version of Stadler or KBCM. I'll stick with those, thank you very much.


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## CuBeOrDiE (Apr 14, 2010)

your welcome


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## guitardude7241 (Apr 14, 2010)

tc is a NUB


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## miniGOINGS (Apr 14, 2010)

CuBeOrDiE said:


> your welcome


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## scottishcuber (Apr 14, 2010)

You cannot preserve the eight edges, you wont be able to do certain moves as they would put the edges in the way of making the centers.
You could make 6 edge pairs instead to allow some freedom, but not eight. This could be a good method, it just needs some tweaking.


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## miniGOINGS (Apr 14, 2010)

scottishcuber said:


> You cannot preserve the eight edges, you wont be able to do certain moves as they would put the edges in the way of making the centers.
> You could make 6 edge pairs instead to allow some freedom, but not eight. This could be a good method, it just needs some tweaking.



He edited his post, explaining that 6 tredges would be preserved. Then he would solve the last 6 tredges and 6 centers. Like I've previously stated, I'll stick with Stadler/KBCM.


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