4x4 last Dedge help.

[andrewvo1324]

What is the algorithm for last pair of dedges are like

XY
YX




?

Thanks.

 

[Crzyazn]

now im beginning to think some people are just lazy....

http://www.bigcubes.com/4x4x4/edges.html

http://www.speedcubing.com/chris/4speedsolve21.html

LOOK AROUND....GOOGLE IS YOUR FRIEND

 

[dChan]

I found this out by myself, geez...

Don't be lazy, experiment! And when experimentation does not work, look around. I was actually trying to figure out how to do the last dedge pair for the 5x5x5 this morning but I did a quick check on bigcubes.com and found my answer.

Ha, yes, google is your friend.

 

[Crzyazn]

Lol aren't they called "tredges" for 5x5?

 

[andrewvo1324]

Yea but is there anything more optimal then L' F U' F U F' L F' L' F (Dd)' F U F' L F' L' F (Dd) or do i suck?

 

[Crzyazn]

Would a site designed by one of the top big cubers in the world steer you wrong? And you can always do some axis turns if something is inconvenient.

But seriously, don't give up on an algorithm so quickly. If you still fail at execution after a week, THEN make another thread like this.

 

[DAE_JA_VOO]

I actually came up with my own algorithm for this case. It's pretty long and almost certainly not the fastest, but it's a very easy one. Let me know if you want it

 

[dChan]

Sorry, I don't know my big cube terms well.

Anyway, all you have to do is understand what the algorithm does and it should be fast. Here is how I solve the last two pairs(it might be a little inefficient but it is fast for me - like around 3-4 seconds maybe):

1. Place one of the tredges at the FL(or FR, it depends on the situation) position and place a solved tredge at the FR position.
2. Place the other unsolved tredge at the UR position.
3. Swap an edge from FL to FR and bring down the tredge from the UR position to FL/FR.
4. Finish by setting up the the three tredges so that you can solve all three at once.

I can't really give you an actual algorithm as I don't remember it as an algorithm but, rather, as a logical operation, however this should give you one idea of how it should work. I don't know if this is inefficient so someone stop me if I am giving the wrong advice(maybe I should just give an algorithm?).

 

[LamentConfiguration]

I hold it:
XY
YX
(on the front face)

then I do a quick flip of one side with a L'FU'LF

then I pair them up with
(Dd)RF'-U-R'F(Dd)'

It isn't one algorithm, but it is pretty short and sweet if I understood what you are asking.

 

[Harry]

I found this out by myself, geez...

Yep, I found this by myself too.

I will give you a hint:
After you align them, you can't replace that with the unsolved one since this is the last edges. Which means, make sure that they are not the last one. If you got what I mean, do that about one time while looking ahead. It should correct it.

Hope you got it!!

 

[rxdeath]

I hold it:
XY
YX
(on the front face)

then I do a quick flip of one side with a L'FU'LF

then I pair them up with
(Dd)RF'-U-R'F(Dd)'

It isn't one algorithm, but it is pretty short and sweet if I understood what you are asking.

this is the bigcubes approach as well i haven't used that exact alg, but our idea is to make it as

X X
Y Y

then pair up (slice the bottom) so its kind of like

X X
Y Y

then you take the X/Y group and flip it upside down

Y X
Y X

then slice back and you are set
Y X
Y X

 

[jackolanternsoup]

as far as i know the most optimal is L2 d2 R U R' F R' F' R d2 L2 (off Dan's site)

 

[Swordsman Kirby]

as far as i know the most optimal is L2 d2 R U R' F R' F' R d2 L2 (off Dan's site)

That's not optimal.

 

[AvGalen]

Common guys, this should be trivial. There are 2 situations possible, horizontal and diagonal swap. And they are both really the same. (Basically the one that's mentioned from Dan's site). I also remember there was an optimal one for situation 1 in 7 moves, but I didn't like/understood that one (you can find it at http://speedcubing.com/chris/4speedsolve21.html just search for R F' U R' F)

Now if we flip the edge in FR, undoing our d layer turn will solve those two edges. To flip simply do the move R F' U R' F. For flipping the edge some people prefer R U2 R' F' U' F but I would definitely suggest the first, shorter alg.

Horizontal swap:
d ("match")
R U' R' (store "matched" pair on U layer)
F' U2 F (put "matched" pair back flipped)
d' (really match)

Diagonal swap:
L2 (or R2 y)
d2 ("match")
R U' R' (store "matched" pair on U layer)
F' U2 F (put "matched" pair back flipped)
d2' (really match)
There is no real need to undo the L2 (or R2 y)

 

[TimMc]

Flip one dedge around so that you've:

YX
YX

This will solve it:
- The algorithm
- Ensuring it works
It's probably not that efficient, meh - it's how I intuitively solved it :-/

Tim.