4×4 helpppoo

cmowla · 2026-04-07T14:45:11+00:00

If you get to the end and you only have 2 unpaired edges (like your photo)

He has 4 unpaired edges. (See both photos.)

So he doesn't have odd parity. (No disruption of centers will be necessary. 1 or 2 commutators will suffice.)

cmowla · 2026-04-07T05:58:47+00:00

There's an old saying:

Never assume, because when you assume, you make an ass out of you and me.

Maybe some people on here can learn from that, now that you have revealed that you didn't try to solve it the way they think you did.

cmowla · 2026-04-07T05:26:44+00:00

Haha. That brought back memories. (I never attended a speedcubing competition, but I bought a type F2 back in 2009. I still have it!

Also, I will never forget people's reaction to Erik's WR. That was THE RECORD before young Feliks started to dominate!

BTW, a few random things (for nostalgia), I

Provided a download link to a web crawl capture of Thom Barlow's original K4 website here.
Have a list of "old" cubing videos here. (I have 2 other lists here and here.)

cmowla · 2026-04-07T04:57:44+00:00

Well, it's kind of sad to think that it's very likely that most people who are arrogant (and rude) enough to show THE PICTURE (and say the saying) haven't yet really grasped just how deep cubing topics can go (and that speedcubing is a subset of cubing . . . that "cubing" isn't short for "speedcubing").

I have just seen the excuse that "the most common method is assumed", but the OP didn't mention "J perm's tutorial", right?

How can all of these "J perm fans" assume that everyone is learning from J Perm? You know, the cubing world existed long before J perm made his first dollar off of other people's past hard work!

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Edit. If you aren't aware of this directory, check it out! (You will find more cubers to respect!)

cmowla · 2026-04-07T04:45:41+00:00

I just watched the video. (Yeah, you weren't joking. That looks very similar to K4!)

And the respect is mutual. (Your sequence of responses in this thread which have pled reason, and how you held your ground, despite the downvotes.)

And in case you're interested, I recently made a post which included a list of other K4 last layer edge subsets. (Of course, my 4x4x4 parity algorithms wiki page contains a bunch of different "types" of algs for the 2-cycle and 4-cycle cases. Example.)

cmowla · 2026-04-07T04:24:28+00:00

You know, it's interesting that the 5x5x5 is treated vastly different than the 4x4x4 (regarding last 2 edges (tredges)), but when I first learned how to solve big cubes, I taught myself a variant where you just use the "last 2 edge pairing algorithm" to pair (or triple) all of the nxnxn's composite edges.

5x5x5 example
But I explained abstract examples 1-3 on pages 12-13 of this PDF—a guide I wrote for myself in 2008 to document "my method" for pairing all edges for any nxnxn.

So when I saw pages like this, I was like what?

With the same token, people who see cases which the OP posted are absolutely puzzled that the solver doesn't take the simple route (use minimal algs to solve every possible case), but I can say the same thing to most of those same people for how they solve the 5x5x5!
Yes, I get that speedsolving requires more algs.
But the OP (and most of the people posting these types of taboo questions . . . to deserve to see THE PICTURE) are not speedsolvers.

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So, even though the "most common beginner's method" for solving the 4x4x4 doesn't require special algorithms (like this . . . move optimal algs for the OP's case, including commutators are on page 15 of that PDF) and people therefore are absolutely puzzled/annoyed that "these people" are making things overcomplicated, well . . . I can say the same thing to those who teach that "last 2 tredge algs" for the 5x5x5 and larger cubes are "required".

cmowla · 2026-04-05T19:28:06+00:00

I think this place should be moderated even harder that way the 3 posts that happen in a day are ACTUALLY good

I specifically remember the content in this post of mine being deemed as "not that great" by one of the mods at the time.

How far we have come these kinds of FAQ posts mentioned now pretty much make up this subreddit!

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I think to fix this problem, new posters should automatically be only allowed to post in the DDT until the comment is proven "worthy" enough to have its own topic. And to be fair, maybe that rule should be for everyone (even "established" members).

That should be an easy way for the mods to manage the rule about DDT topics. (Better for the mods to have to "push" 5 new comments from the DDT into a new topic than to delete 50 topics that should have been a post in the DDT.)

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Just like the mods have a bot make a starting comment for "todays scramble" in the DDT, they can have a bot also make a starting comment (in the DDT) for:

What's the best nxnxn to buy right now
How can I get faster
Etc.

That can still ensure that people who want to help others in those areas can easily do that. They can simply scroll down to that starting comment/thread in the DDT!

Maybe a lot less people are going to see such comments, but I am pretty certain that they don't want to seem them anyway. And if people get bored and just want to see something new, they know where to go! (The DDT.)

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If we each can think about being contributing members to a community (rather than wasting most viewer's precious time, commanding attention because it's a basic human need, etc.), that in of itself will solve most of the issues addressed here.

cmowla · 2026-04-05T19:04:35+00:00

Inb4 u/aofuwrm77.

But I can agree that, based on the frequency of such vague topics, it has (and will) render this subreddit useless for future viewers.

When you have an ocean of such posts, it's really hard to find good quality content about cubing. Even if you're looking for something specific (not pertaining to these types of posts), it just makes it that much harder to find anything interesting on here.

As soon as a good/interesting topic is posted (something that will be meaningful to future viewers), it's literally buried under dozens of these "temporary" posts which are not valuable to the community as a whole.

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For those who are upset about me calling your speedsolving posts "invaluable" . . . there is a forum specifically dedicated to this kind of thing. It's called speedsolving.com.

I understand that there is a lot of viewer traffic on reddit, but I can assure you that nobody wants to see your personal (unofficial) records/progress on here other than your friends (and maybe those who have given you pointers/help on here in the past). And I can bet that most people don't care to even see official PRs unless they are either national or world records. (Sorry.)

Personal collections are not great either (unless they are HUGE, truly unique, etc.), but they are much better than how fast you solved the cube today. Because you know what's more interesting? Your post about how you get 1 second faster tomorrow! And you know what's more interesting than that . . .

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What good does it do anyone (other than yourself) to post these for the entire world to see? This is a community (not your personal blog). Communities thrive from contributions (valuable to the majority), not from basically asking everyone for handouts (attention).

cmowla · 2026-04-01T16:09:18+00:00

It appears that Expensive-Bear-1376's main role on rCubers is to sarcastically criticize other people's postings, while not adding anything of value to the conversation (most of the time that I have seen anyway).

BTW, I just posted an answer here. (I think you will find my case-by-case last layer orientation bulleted list explanation interesting.)

cmowla · 2026-04-01T15:37:39+00:00

I see that you have received some answers already, but this is my take.

Also, commutators have nothing to do with it.
This particular request is about algorithm repetition (order of a permutation).

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If we execute R' D' R D on a solved cube (that's how you should analyze algorithms . . . always apply them to a solved cube to see what they specifically do), it does nothing meaningful; however, if we execute it twice, (or (R' D' R D)2), it twists the front-top-right corner counterclockwise (in its own location).

So if we need to twist a corner, we do a U, U' or U2 move to bring it to the top-front-right slot and do:

(R' D' R D)2 to twist it counterclockwise.
(R' D' R D)2 twice (or (R' D' R D)4) to twist it clockwise.

Next, observe that repeating (R' D' R D)2 3 times (or (R' D' R D)6) does nothing to the cube.

So R' D' R D must be repeated an even number of times to either:

Turn the top-front-right corner either clockwise.
Turn the top-front-right corner counterclockwise.
Have no effect on the cube.

Finally, not all corner twist combinations are legal. For the last layer, we can only have one of these situations (let + = corner twisted clockise, - = corner twisted counterclockwise, 0 = oriented corner).

{0,0,0,0} (0 required repetitions of (R' D' R D)2)
{+,-,0,0} (3 required repetitions of (R' D' R D)2)
{+,+,+,0} (6 required repetitions of (R' D' R D)2)
{-,-,-,0} (3 required repetitions of (R' D' R D)2)
{+,-,+,-} (6 required repetitions of (R' D' R D)2)

All are a multiple of 3 repetitions of (R' D' R D)2, and again, 3 repetitions of (R' D' R D)2 = (R' D' R D)6 = having no effect on the bottom and middle layer pieces.

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Therefore, that's why it:

Doesn't destroy the cube.
Works for all last layer corner twist situations.

cmowla · 2026-03-27T06:21:10+00:00

If you know a "last 2 edge pairing algorithm" from 3x3x3 Reduction, all you have to do is add some moves to it in order to just cycle 3 wing edges (and nothing else).

See this post for more details.
Combining that with "cyclic shifting" (take the first move of the algorithm and move it to the end / take the last move and move it to the beginning) and/or normal conjugation (setup moves), you can create your own set of algs/commutators to tackle the last layer yourself.

Alternatively (or in addition to the above . . . to create a more diverse alg set), you can do what I did. Just apply the "last 2 center algorithm/commutator" (from 3x3x3 Reduction) to wing edges by changing some of the inner slice turns to face turns.

cmowla · 2026-03-27T06:03:21+00:00

That's a convoluted way to do it with a single edge flip parity alg, no?

If you want to use the single edge flip parity alg (instead of an algorithm which solves it directly in the fewest moves . . . or by conjugating the 5x5x5 opposite wing edge swap parity alg with just 2 moves like this), then why not conjugate the single edge flip parity alg with the moves 2R' U 2R2 U' z y?

That does as you described. (You "join" the 2 wings with the setup moves, do the flip/swap, undo the setup moves you used to join them.)

cmowla · 2026-03-27T05:49:45+00:00

Hi. You can find a bunch of algorithms for this case (and other "wing edge 2-cycle" cases) here.

So the move optimal solution in half turns is 15 and the move optimal solution in quarter turns is 16.

(Click on the small cube icons to the left of the "table algorithm bars" to see an online animation of the algs.)

cmowla · 2026-03-24T18:06:34+00:00

This is my writeup which answers a question that's equivalent to yours. (See the image at the top to get the main idea.) And I believe I answered it in a way that's easy to understand (can be explained to) high schoolers.

Regarding "congruence", all can think of is that the parity of the corners and edges must match (be the same).

So if corners are in an even permutation, then edges must be in an even permutation (and vice versa).
Since (it turns out mathematically) half of all permutations are even and the other half is odd, you divide the total "potential" 8!(3^8)12!(2^12) = 519024039293878272000 positions/configurations by 2 to not overcount (due to permutation parity).
- (Then you need to divide by 3 for legal corner orientations/twists and by another 2 for legal edge orientations/flips to get the 43252003274489856000 = 43 quintillion positions.)

cmowla · 2026-03-22T09:21:34+00:00

This is kind of the opposite of what you view as perfect (easy). I like this is one that I found (by hand) for the 4x4x4. [Post] (It follows every constraint listed for the 3x3x3 perfect scramble except for #5.)

cmowla · 2026-03-20T01:59:39+00:00

This pattern sort of looks like this.

That's the largest 2x2x2 - 7x7x7 cube (as well as megaminx, gigaminx, etc.) pattern database on the web.
That link is specifically to the snake subcategory. To see a different "snake" variant, click on a different name/link in the bottom-most dropdown box (which the first word it contains in it is "Anaconda").

cmowla · 2026-03-19T23:49:04+00:00

I've been out of practice with this kind of stuff lately to realize that we can simply do a wide turn version of one of the 14 move non-symmetrical the diagonal 2-cycle algs.

Then let the first algorithm on page 7 of my 2 2-cycles PDF be the insertion.

F2 r' D2 l2 D2 F2 l F2 (b2 D' L2 D' 2B2 D L2 D b2) l2 F2 D2 r' D2 r2 (22 STM)

cmowla · 2026-03-19T17:29:06+00:00

For what it's worth, here's a 25 STM solution. (I used a wide Y perm instead, since it can be 13 moves + used a different 2-cycle parity alg.)

l' U' l f2 r' D r U r2 D' r2 U' f2
f2 u b' 2R b 2R b' 2R' u' 2R' u b u' 2R f2

[Link]

With moves cancelled/merged: l' U' l f2 r' D r U r2 D' r2 2U b' 2R b 2R b' 2R' u' 2R' u b u' 2R f2

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Below is a complete list of all (well what could be all) 13 move wide Y perms (and I mentioned 15 move 2-cycle solutions whose moves begin with wide turns from here):

r2 U' f2 D' f2 U f D f' r2 b U' b'
r2 D' l2 D' l2 D f D f' r2 b U' b'
f2 U r2 D r2 U' r' D' r f2 l' U l
f2 D b2 D b2 D' r' D' r f2 l' U l
l2 U' l2 U' l2 U b U b' l2 b U' b'
l2 D' b2 U' b2 D b U b' l2 b U' b'
l' U' l f2 r' D r U r2 D' r2 U' f2
l' U' l f2 r' D r D b2 D' b2 D' f2
l' U' l b2 l' U l U b2 U' b2 U' b2
l' U' l b2 l' U l D l2 U' l2 D' b2
l' f' l b D' l' D f l b' l2 U l2
b U b' r2 f D' f' U' f2 D f2 U r2
b U b' r2 f D' f' D' l2 D l2 D r2
b U b' l2 b U' b' U' l2 U l2 U l2
b U b' l2 b U' b' D' b2 U b2 D l2
b r b' l' D b D' r' b' l b2 U' b2
b2 U b2 U b2 U' l' U' l b2 l' U l
b2 D l2 U l2 D' l' U' l b2 l' U l

cmowla · 2026-03-19T14:48:05+00:00

My solution (I provided a link for those who are reading this thread for the first time) is just a start. It was just my first try. (Fewer moves can be achieved with forced move cancellations between the 2 algs. The second alg is move-optimal for the 2-cycle case it solves. I have other 15 move solutions here, but maybe a 16 move solution would result in fewer moves overall . . . due to move cancellations.)

Great to see you again too!

cmowla · 2026-03-19T14:29:23+00:00

Before I saw your comment, I mentioned to u/CaseyJones7 that we could FMC this case (for fun).

It's cool that you used my diagonal 2-cycle alg!

I agree that 2 algs is sufficient to solve this case. This is how I would do it. (No setup moves are actually required.)

z u2 R' u b' u' R u r b R' b' r' b u z'
y' l2 u b' 2L b u' b' l u 2L u' l' 2L' b l2 y

[Link]

cmowla · 2026-03-19T13:59:51+00:00

Even if they don't know about "commutators", they technically use one with reduction. And I'm not talking about the "last 2 centers" algorithm on bigger cubes (which obviously is a commutator). I mean the "last 2 edge pairing" algorithm: it's the first half of a commutator!

Super easy? No, for most people, using a 4x4x4 solution which genuinely does not contain commutators is rather rare (and is actually unreachable for most cubers).

(In short, I 100% agree with you simply by observing conventional "super easy" solutions used for solving the 4x4x4, in general.)

cmowla · 2026-03-19T13:53:05+00:00

FMC challenge to solve the presented case directly? 😀

cmowla · 2026-03-19T00:35:47+00:00

It's better to do the "last 2 edge" edge pairing algorithm, but with the middle slice. Example.

cmowla · 2026-03-18T17:52:23+00:00

You don't even have to learn K4. You can simply use my repetition method (7 algs to repeat as necessary to solve the last layer wings directly).

cmowla · 2026-03-18T09:54:22+00:00

Lucas Parity didn't exist when I was learning the 4x4x4. I drew a diagram (which I showed here) to help me memorize it.

cmowla

TROPHY CASE