I understand this thread made most of you angry because this topic has been bled out enough by beginners, but please stick with me.
To be honest - I'm not attempting to become a speed cuber, so learning the algorithms is really no fun for me. I want to enjoy the process, not the time spent solving. I've learned to solve the 3x3 on my own, and I'm not using any algorithms per-say, just some that I came across on my own.
Now with the 4x4 parity case where one edge is incorrectly flipped, I'm trying to come up with a solution that fixes it, but I really don't mind the solution scrambling the rest of the cube. The algorithms are there to keep everything in place - I don't need that.
Could anyone point me towards a solution that fixes the flipped edge in an intuitive way, and if it involves messing up the cube - I don't care. I tried breaking up the flipped edge just like you would pair an edge up, then swap the newly created edge that has one piece of the flipped edge with an already so. So the exact same steps that are used to create on edge, just the end result is that I get a broken flipped edge, along with 2 other edges that get messed up. So then I reassemble the edges and the incorrect one from the beginning get's flipped correctly, but another edge flips and all I'm end up with is changing which edge is flipped.... A friend of mine told me that this method works on the 5x5, but it seems that on the 4x4 it's not working, I'm doing something wrong..
I hope you all won't be too mad with this question, and thanks for any pointers
The incorrecly flipped edge case is caused by the 24 wing edges having odd permutation parity. The only way to correct odd permutation parity is to perform an odd permutation on that piece orbit. Doing so will give the 24 wing pieces even permutation parity, and that will "unflip" your edge group.
The only type of turn that performs an odd permutation on the wing edges is an inner-layer quarter turn. More generally an odd number of inner-layer quarter turns will perform an odd permutation on the wing edges. What this means is that you will be able to assemble your edge groups (dedges) such that there will not be one flipped incorrectly at the end of your solve.
If you are going to come up with a process to flip that one edge group correctly, then it must contain an odd number of inner layer quarter turns.
Before someone else just posts an alg that does this, I'll give you a hint that will point you down the right path to figuring out why the "intuitive" algs for OLL parity work.
The hint is: don't focus on the edges/breaking up and repairing up edge groups. This strategy uses something called "conjugate" maneuvers, which in this specific case will always do an even number of inner-layer quarter turns (question: why won't this fix your flipped edge?). Try instead to focus on breaking up the centers, then restoring them. The key here is to somehow break up and rebuild the centers, but using an odd number of inner-layer quarter turns to do so.
I think that if you break up the centers and rebuild them using an ODD number of inner layer quarter turns, that should get rid of parity.
I'm not 100% sure tough
There is a very intuitive way I came up with to deal with this.
I know you said you didn't want algorithms, but just look at how my algorithm sets up the centers: you can set up the centers like this with whichever moves you want without memorizing or even looking at these moves I have provided: f2 u' r2 Uw2 S' (look at how the red and orange center pieces are setup in the right inner layer slice).
[1] Do whatever moves you want to get the previous setup for the centers.
[2] Using only 3x3x3 moves, insert wing edges (4x4x4 individual edge cubies) into that slice so that you have two same two color edges opposite (diagonal) to each other in that slice. Make an effort to preserve the center setup you did.
For example, F d' S r F Rw2 f2 u' r2 Uw2 S' (notice that I made the same formation for the yellow and white centers in the r inner layer slice, but now I have also put in the blue-white and green-white edge cubies in that slice opposite to each other, respectively.
From this point, all you need to do is a PLL parity algorithm (which I assume you consider to be easy, but if not, let us know and a lot of people can explain different ways of looking/dealing with it which should make it understandable).
I have studied the 4x4x4 parity algorithm domain (see my signature for links to the wiki page and to my methods thread), and I have studied how different algorithms (approaches to fixing parity) in different move sets, and I can tell you that this is by far the easiest way to deal with parity. I don't particularly like the length of the algorithms, but we're not worrying about that. For those who are curious to see if there are "prettier" setups to the centers, here's a setup for opp. double parity: [Rw Uw2 U' Rw L2 U2 Lw' U' Lw R F2 U' R' U: r]
and it modified a little: [Rw Uw2 U' Rw L2 U2 Lw' U' Rw R U2 (Lw r') U' R2 U: r]
And, you are not limited to just two specific dedges, so I found a shorter algorithm for adj. double parity, for example. [Lw' Uw2 U Lw' R2 U2 Rw U Lw' U2 F' R2 F: l]
I've seem to released the Kraken with my question and use of certain words.. In my defense I'm just acknowledging that the "fix parity without long algorithms" question has been a focal point of beginners trying to opt out of learning long algorithms and I'm sorry if in some way I have offended you, seriously, I am. As I said in the first post, I'm not looking for a easy solution, in fact, I'm looking for a longer solution, a solution that I can understand what I'm doing when turning each and every face, instead of knowing the steps in order to reach a certain expected result. To me - solving the cubes no matter the size of it, is fun because of the unpredictability of them. I want to be able to SOLVE a cube, not learn the steps necessary to do so.
@cmhardw - I will try the odd inside quarter turns and see how I fair. Thanks to you and others who didn't just jump into a flame war.
I started with a flipped edge, did that algorithm, ended up splitting the edge pieces opposite each other, then combined them with Dw R F ' U R ' F Dw, got the cube scrambled of course but the edges and centers where in tact, just had to reposition everything. And I ended up fixing the parity error So thanks for this algorithm, I can wrap my head around it and understand what it does and how it does it.
Another thing I found, maybe most of you already know this, but if anyone stumbles upon this topic with the same goal as I have - they'll be pleased to know there is an easy(ier) way to fix adjacent edges being switched (by easy(ier) I mean more comprehensible, at least to me..) . What I did was swap 3 centers using R U' R U R U R U' R' U' R2 and what I got was two edges opposite each other that had to be swapped, and by using r2 U2 r2 Uw2 r2 u2 I swapped them and fixed the adjacent parity error. I understand fixing adjacent parities involves using the same algorithm used to swap opposite edges, just adding R' U R U' in front and U R' U' R at the end, but to me the way I found is more "user friendly" in terms of understanding the steps involved.
Also maybe someone knows if these same methods apply to bigger cubes? I'm particularly interested in knowing if this could work on the 5x5.
Thanks for everyone who helped, you're awesome!!
Edit: Also, I understand this forum, website and wiki are all leaning towards speed solving, but does anyone know where to find algorithms or simply theory involved in certain situations, just like I had with this parity error. It was awesome that cmhardw explained why the parity errors occur, and hinting a possible solution. I tried it but failed, I'm just new to the cubes, but cmowla got my right back on track and now I can wrap my head around the parity errors and how to fix them. I would love to find a resource that would just provide theory or basic steps needed to solve certain problems, instead of having a bunch of algorithms with no explanation how they work.
The incorrecly flipped edge case is caused by the 24 wing edges having odd permutation parity. The only way to correct odd permutation parity is to perform an odd permutation on that piece orbit. Doing so will give the 24 wing pieces even permutation parity, and that will "unflip" your edge group.
The only type of turn that performs an odd permutation on the wing edges is an inner-layer quarter turn. More generally an odd number of inner-layer quarter turns will perform an odd permutation on the wing edges. What this means is that you will be able to assemble your edge groups (dedges) such that there will not be one flipped incorrectly at the end of your solve.
If you are going to come up with a process to flip that one edge group correctly, then it must contain an odd number of inner layer quarter turns.
That is exactly how I did solve the 4x4 for the first time, from a inner layer turn and start correcting the cube again will take 20+ turns , but it is quite intuitive
In general you can solve any other puzzle with odd permutation like this.... not the fastest way but intuitive
Great! No need to remember algorithms for each cube and I don't mind resolving Also I found that this way I always end up with a parity error involving swapped edge pieces either adjacent or opposite. But that's easy to fix so I'm not complaining.
Hmm, the standard Rw (U2 Rw)4 has always been understandable to me as it is, but only now did I realize you actually get the same by doing the centers slowly one by one as
Rw (U2 Rw U2 Rw') (Rw2 U2 Rw2) (Rw' U2 Rw), after cancellations it's the exact same algorithm (not that surprising, but I had never thought about it). And re-solving centers one by one like that should be possible for anyone.
I don't think I've seen cmowla's setup f2 u' r2 Uw2 S' before. Very nice. Back when I created Uw2 Rw2 U2 r2 U2 Rw2 U2w of course I also tried it for the other parity, but couldn't find a short setup.
I don't think I've seen cmowla's setup f2 u' r2 Uw2 S' before. Very nice. Back when I created Uw2 Rw2 U2 r2 U2 Rw2 U2w of course I also tried it for the other parity, but couldn't find a short setup.
Thanks. I'm surprised you haven't seen it, as I posted it more than a year and a half ago, here. (I think the Latex has a conflict with the mobile device settings for SS, so that's why that thread has a bunch of lines of error code at the beginning).
I can't remember all of my favorite move count break-throughs which I am unsure if you have seen them, but these were the ones which I thought of first
Up to this point, I posted most of the algorithms I found on the 4x4x4 wiki page (except for some algorithms I found recently or algorithms I found which are for cases other than what is on that page), so go check it out. I have added images recently for difference cases, attributed algorithms to their original founders (but I left a lot of algorithms' authors just be "author" because I wasn't sure who found it first), and I have added move counts (I created a variation of the alg template and used it on that page).
Hmm, the standard Rw (U2 Rw)4 has always been understandable to me as it is, but only now did I realize you actually get the same by doing the centers slowly one by one as
Rw (U2 Rw U2 Rw') (Rw2 U2 Rw2) (Rw' U2 Rw), after cancellations it's the exact same algorithm (not that surprising, but I had never thought about it). And re-solving centers one by one like that should be possible for anyone.
I don't think my algorithm itself [f2 u' r2 Uw2 S': r] is as easy to understand as the setup it creates: it sets up the centers so that if you do a quarter turn to it, no visual change is made to the centers (as you would know more than anyone else).
(r U2)4 r is an algorithm one must see executed to understand it (and on a small level at that, assuming you don't do any type of arithmetic such as your decomposition or mine), but CodilX was asking for a concept rather than getting used to an algorithm, so I believe my approach was more appropriate for this request.
On a side note, the conjugate [r2 d' r2 Uw2 S': r'] affects the wings in the same manner as (r U2)4 r, as I mentioned at the very beginning of the 4x4x4 parity algorithms wiki page.
For example, on the 6x6x6, inner orbit, outer orbit (this is in SiGN notation, but you can see the algorithm executed in the cube animation).
I'm surprised you didn't look at the setup to preserve the dedges, as it's the same setup except that, after you set up the centers this way, you use 3x3x3 moves to place in individual edge cubies into the slice to flip a dedge and swap that dedge with another dedge.
Anyway, a much prettier algorithm I found about a year ago which simply accomplishes this task of inducing (or breaking) an odd permutation in the wing edge orbits which the extra quarter turn runs through for all even cubes is: [Rw' Uw2 U Rw' U2 Lw U Rw' U2: r].
About 6 months after I found f2 u' r2 Uw2 S', I went back and created a setup which works on all size cubes, odd and even, but it's longer. In fact, this is how I found the "prettier" setup algorithms. (This again sets up the r slice's centers): Rw' Uw2 Lw' Rw' Uw2 Lw2 U2 Lw' U' Rw' U2
Notice that the above setup also sets up both the l and r slices complete on even cubes. To set up both the l and r slices on odd and even cubes, we only need to modify the ending a little: Rw' Uw2 Lw' Rw' Uw2 Lw2 U2 Lw' U' Lw' Rw x' B Rw' U2 x
(Obviously the algorithm above can simplify if applied only to even cubes or to odd cubes, but the above translation works for all size cubes).
Edit: Also, I understand this forum, website and wiki are all leaning towards speed solving, but does anyone know where to find algorithms or simply theory involved in certain situations, just like I had with this parity error. It was awesome that cmhardw explained why the parity errors occur, and hinting a possible solution. I tried it but failed, I'm just new to the cubes, but cmowla got my right back on track and now I can wrap my head around the parity errors and how to fix them. I would love to find a resource that would just provide theory or basic steps needed to solve certain problems, instead of having a bunch of algorithms with no explanation how they work.
If you are interested in parity algorithm theory, I humbly believe I am practically the founder of it.
Now that I have provided a way for you to deal with parity without algorithms, I introduce you to algorithm theory (I could probably easily write a 200 page book on it, but when I first mentioned writing and selling a book to people on this website, I got negative responses...I wonder if things have changed since then? Yes I have provided many posts on this subject, but I didn't explicitly write how I thought of everything and how I mentally link things together to be able to come up with all of these short algorithms without a computer solver...). If you have the patience to read through some of my posts, you will see me break down very "complicated" algorithms into simple pieces in my "derivations". Many of the algorithms I made derivation posts for I am the original founder for.
I have created a PDF document (although I have some unwanted duplicates in it, some decompositions in it are "incorrect" and it's incomplete...but most of it is okay) which contains decompositions of various parity algorithms (I use a different notation in that document than what is on alg.garron.us, but I provided details how to view them using cubeTwister, in the same post which I posted this PDF to),
I have written posts like this from time to time in other threads besides my main two.
If there is a specific parity algorithm(s) you would like an explanation for, feel free to ask, but I most likely have already touched on it in the links provided above. I do have derivations for common parity algorithms in video form (on YouTube), but I hope to make more videos in the future when I have time.
(r U2)4 r is an algorithm one must see executed to understand it (and on a small level at that, assuming you don't do any type of arithmetic such as your decomposition or mine), but CodilX was asking for a concept rather than getting used to an algorithm, so I believe my approach was more appropriate for this request.
Well, I think one can easily come up with the version I showed where the centers are slowly re-solved one by one. And there's no magical setup like in yours. I wouldn't call mine a "decomposition", btw. That sounds like I started with (r U2)4 r and analyzed it. I didn't. I did the one-by-one centers solution, and *then* noticed that after cancellations it's the same as the compact version.
)
So thanks for this algorithm, I can wrap my head around it and understand what it does and how it does it.
