9/12 dedges solved, 3 left... Where should I go from here? (4x4x4)

[StringStrider]

Hey guys, me again with another noob question...

I've searched all over for how to deal with creating the remaining dedges if there are 9/12 (3 left) solved, I can't seem to figure out on my own what to do, because it seems everything I do leaves me solving one, and breaking apart a previously solved dedge... :confused:

Most places show me the 6 pair or 2 pair methods, but none of them show me how to deal with 3 left over? :fp I feel dumb... Is it some sort of midway parity I have overlooked?

I've tried orienting them to "solve" 2 and have 1 left which would solve itself on its own (or leave me with a parity I could search an alg for with a final dedge with swapped sides) but that ends up with screwed up centers no matter how hard I try...

Advice?

lol. If only 4x4x4 was as easy as 3x3x3

PS: I tried looking for videos, but for some reason youtube doesn't work for me at home or work, so unfortunately a link to a youtube video most likely won't help

 

[mcciff2112]

I don't want to give too much away, because I think you should do it on your own, but if you really want it:

Spoiler

You need to replace one of the edges you've just solved with another unsolved edge, just like you've been doing for the edges so far. If you leave one of the three unsolved edges in one of the solved layers (the layers with all of your already solved edges in it), you can slice to solve one of the other unsolved edges, then replace that with the unsolved one you preserved earlier. Depending on the way you insert the preserved unsolved edge, you will either finish the edges or have two left in which you will need to swap a piece on each one. I would take a good look at which way you're planning on inserting that edge before you actually do it to avoid this case.

If you do end up with this "parity":

Spoiler

you can fix it easily by slicing, the performing an edge flipping algorithm. Hold it so the two unsolved edges are in the front-right and front-left positions, the colors should match horizontally(Meaning it should look something like:
0000
X00X
Y00Y
0000
X and Y being the two colors, and 0's being irrelevant pieces ).

Now do the algorithm Dw R U R' F R' F' R Dw'

Fin

 

[Innocence]

4x4x4 is as easy as 3x3x3, just longer with more steps! ^_^ If you know how to solve 2 dedges, then the third will be on its own, therefore have no other way to be other than solved. I don't really get the problem.