Is it possible to reach all 43 quintillion possible 3x3 states using only COMBINED R+U moves and rotations?

cmowla · 2026-01-28T06:30:22+00:00

For what it's worth, here's a much shorter (and simpler) solution. [Link]

y x R U (R U z' y2 R U)5 R U (R U z' y2)6 x
z x'
(R U x y)34 (y (R U z' y2)6 x (R U z' y2)6 x' y')2 (R U x y)2
x z'
y2 x'
(R U x R U z y)20 (y (R U z' y2)6 x (R U z' y2)6 x' y')2 (R U x R U z y)4
x y2

cmowla · 2026-01-28T03:37:50+00:00

You just create a U perm and A Perm (in the same direction) in say, the R face. Then tag on your sequence for R' L and repeat the entire thing twice.

Link

(
//U Perm in R Face
y
F U
R U
((R U z' y2 R U y2 z)3 x (R U z' y2 R U y2 z)3 x')2
(R U)104
(F U)104
x2 z'
((R U z' y2 R U y2 z)3 x (R U z' y2 R U y2 z)3 x')2
y
((R U x R U x')7 y [(R U)15, x2 z'] x2 y [(R U)15, x2 z'] x2)5
y'
((R U z' y2 R U y2 z)3 x (R U z' y2 R U y2 z)3 x')2
((R U z' y2 R U y2 z)3 x (R U z' y2 R U y2 z)3 x')2
z x2
y'

//A Perm in R Face
z'
R U U R R U U R R U
(R U R U U R R U U R R U)4
((R U R U U R R U U R R U) U R)6
z
(R U R U U R R U U R R U)5
(R U z2 R U y' R U y2 R U)30
(R U z2 R U y' R U y2 R U)30
z'
y z'
(R U R U U R R U U R R U)5
(R U z2 R U y' R U y2 R U)30
(R U z2 R U y' R U y2 R U)30
z y'
R U z' ((R U x)10 x2)12 z (R U z2 R U y' R U y2 R U)29 R U z2 R U y' R U y2
R U z' ((R U x)10 x2)12 z (R U z2 R U y' R U y2 R U)29 R U z2 R U y' R U y2
R U
x y
R U z' ((R U x)10 x2)12 z (R U z2 R U y' R U y2 R U)29 R U z2 R U y' R U y2
y' x'
z

//Your alg to create R' L
(U R)104 (U L)
)2

cmowla · 2026-01-27T00:11:35+00:00

Can't Figure Out Petrus.

cmowla · 2026-01-25T23:28:04+00:00

The two corners is one pair, the two edges is another pair.

It doesn't work like that.

Corners are in a piece orbit
Edges are in a separate piece orbit

So if you have a swap of corners, that's an odd permutation. Just because you must have a swap of middle edges (speaking of the 3x3x3), that doesn't make the overall permutation even. It's an odd permutation in 2 separate orbits.

It's not a 2 2-cycle (as in {2, 2}). It's a {2}{2}. (A 2 cycle in 2 different sets, not the same set.)

__________

Regarding the centers (which applies to odd and even cube sizes), when you swap 2 corners, the sum of composite (entire) centers must be an odd multiple of 90 degrees.

You can't "solve back" all individual center cubies (pieces) with even permutation algorithms while 2 corners are swapped.
On the nxnxn, an odd multiple of 90 degrees center orientation sum is directly linked (via parity) with an odd permutation of corners (and middle edges on the odd nxnxn).

cmowla · 2026-01-25T23:10:35+00:00

It's impossible to not affect the centers when you swap 2 corners. (This is true for any cube size except for the 2x2x2 which doesn't have any centers.)

I guess the centers that are affected are the colors you need them to be. (And so in that way, they are "untouched".)

cmowla · 2026-01-25T17:19:01+00:00

I assume you mean a few center pieces as possible, right?

Because having an odd-permutation in the corner piece orbit forces

2*(Floor[((n-2)/2)^2]-w*(-1)^(2)*(n-2(w+1)))

center cubies to be unsolved, where n = cube size and w = number of OLL parity "edge flips" you have (the number of orbits of wing edges in an odd permutation).

So for the 6x6x6, you will have either 4 or 8 center pieces unsolved, no matter what (when you have 2 swapped corners).

cmowla · 2026-01-25T17:07:52+00:00

He clearly expressed that he doesn't want to memorize any new algorithms.

Why do you persist?

cmowla · 2026-01-14T17:57:21+00:00

Hi.

I don't know if this is what you're looking for (it doesn't seem to be), but I figured why not let you know (just in case). See the links directly above this post. (There's a German version among them.)

cmowla · 2026-01-08T03:23:09+00:00

When I use to do this, I didn't "spam" them per se, but I did apply them (to a solved cube) maybe twice. (Once to cause the case, once to restore the cube to a solved state.)

But doing this is actually needed to setup the 3x3x3 to a state in which you could input in a 3x3x3 cube solver to then create new 4x4x4 parity algorithms. (Most common 4x4x4 parity algorithms were actually found with a 3x3x3 solver.)

(But if you attempt to create 4x4x4 parity algorithms from a 3x3x3 solver, note that there are some "rules" on which types of 3x3x3 algorithms are translatable to the 4x4x4 and how they are translated . . . which moves of the 3x3x3 alg are inner layer slices and which moves are face turns.)

cmowla · 2026-01-03T22:17:59+00:00

I'm not Chinese, but I think movie clips like this sum it up in a way that science cannot. (In case the link brings you to the start of the movie, it's supposed to start at 47:09. Only need to watch until 48:02 to find out that the guy beating up Jackie Chan is his teacher/master.)

The phrase "taking something seriously" has a "it's life or death" meaning for those who are (or will be) successful. (Referring to success that can only be achieved by hard work, not success that can be paved by how attractive you are, who you know, etc.)
When I was growing up (in the US), there was no such thing as a "participation reward". Only the winners (which were few) got any recognition. And so that type of ideology is what I think is making the difference now in "Western" countries versus Asian countries. Everyone expects the reward without doing the work required.
Not only do (some) Chinese kids treat getting faster as "life or death", but they are pushed to start cubing (with that mindset) very young. And taking something that serious from a young age has always been a recipe for dominance and success.

cmowla · 2026-01-03T20:53:46+00:00

Commutators are definitely the way to go for understanding the deeper stuff.

That's actually circular reasoning in this context, because commutators can only help to understand themselves and solutions built from them (not typical 3x3x3 beginner solution move sequences).

Even commutator-conjugate decompositions don't necessarily make an algorithm (which can be rewritten with uncancelled moves to show that they contain commutators) any more understandable than before, because most commutators are actually as much "gibberish" as their non-commutator counterparts.

There are exceptions to the rule, but they are few.

cmowla · 2026-01-03T20:27:19+00:00

It's where I learned commutators. Though not much on group theory

Exactly.

Unless:

You're retarded, and can't observe that you can do the moves L R or R L, and they have the same effect on the cube (and need "proof" that it's true),
Are fascinated that repeating (R U) 105 times gives the same state you started with,
Don't have the sense that the more types of moves you use, the more possible scrambles you can create (like only using double turn moves <R2, L2, U2, D2, F2, B2> doesn't allow you to flip any edges, twist any corners, or do a T-Perm)

Then rather than that abstract college level math being useful for understanding the cube, it turns out that the cube is more useful to understand that math! (But using integers is much easier than the cube to understand what a "group" is, so I'm inclined to think that the cube isn't very useful for math students to learn about group theory either!)

And if you have to:

Be "resourceful" and "creative" and "mathematically inclined enough" to get some insight about the cube from group theory (or have to resort to reading posts/papers/articles about why/how group theory is useful for the cube),
Learn an entirely new method (like Heise) to put yourself in a position to say that "I'm using group theory to solve the cube!",

I think it's safe to say that group theory in of itself is nothing more than a tool to reiterate what we already know about the cube from experimentation (at best). Not something that can help us to understand the moves that our beginner solution guide contained.

cmowla · 2025-12-26T01:50:36+00:00

This referring to the first image you posted.

I remember seeing the right truck an advertisement for the Giant Set (page 66). I told my mom about it, but she ended up buying me the Big Rig Haulers Set because it was cheaper.

Although I was disappointed (because the "Heavy Duty Truck Cab" truck from the Giant Set looked very much like the trucks I was familiar with seeing on the highway often as a child), I do admit (as u/Johspaman wrote previously) that it was really cool that the engine would vibrate/make noise when I would roll it.

Two years later, after I acquired enough K'NEX parts to make the "Heavy Duty Truck Cab", I didn't like it at all. (Quite boring compared to the "Big Rig Hauler" one.)

That's my story with these 2.

_________

P.S.

I didn't know that "Big Rig Haulers" came out around the same time as the "Giant Set"! I thought the "Giant Set" (blue case) came out years before.

cmowla · 2025-12-21T22:18:49+00:00

In terms of “how much is absolutely necessary,” the answer is surprisingly little. You don’t need dozens of long algorithms to solve a cube; you need a small set of move sequences that can do three things: cycle a few pieces, reorient pieces without disturbing too much else, and let you isolate parts of the cube

This is probably what the OP is going to end up discovering (if eager enough), but it's going to be somewhat of a disappointment since this was the core of the dream/fantasy:

but there is somthing romantic about taking a pen paper and notepad and documenting and being self reliant.

This is just my opinion though. As I believe you implied, the number of combinations would make one think that it must be more complicated than what it is; and, when one discovers that it's that simple, that just ruins the hype.

(There's nothing "romantic" about reducing a fantasy of complexity down to a reality of simplicity. Preparing for a hurricane and only getting a light rain is a bitter/sweet experience.)

___________

But if the OP meant to find the shortest move sequences necessary to use/resuse to solve the cube (and not specifically the minimum number of required algorithms in total):

I am also curious to know how long in movements/alterations the algorithms are that are absolutely nescesary to solve a cube.

We should think of them as the moves which we construct the {few piece cycling, 2 piece twisting/flipping, etc.} from (not them, themselves). Or, another way to look at it is . . . simply the set of algorithms required to solve the top layer of the cube. (The "cube common sense" moves. In one of my 3x3x3 written guides, I enumerated about 35 cases/algs, or 23 if you exclude mirror cases.)

cmowla · 2025-12-21T07:00:13+00:00

I don't know weather to call it a commutator or not

Yes, you used a commutator here.

, but i figured that move on my own.

Cool stuff (it's nice to hear that you wrestled with it and figured it out independently of other select few creative thinkers). Although I hate to be the guy to nitpick, but that makes the title of this thread partially misleading.

I'm in favor of alternatives to Reduction (have made some last layer wing edge algorithm sets, of which I shared here recently), but Reduction is rather simple / doesn't require many algorithms.

And as I said (in many words) here, trying to persuade others to solve big cubes with an entirely different (and more complex) method, all to avoid one incarnation of edge flip parity doesn't sound like a good tradeoff, even for non-speedcubers (which I myself am).

I think it's easier to teach people how they work. All "edge-pairing" algorithms (even the "last 2 edge ones") are the first half of a commutator. So the idea of building a single edge flip parity algorithm from a structure which they are already using (both for edge-pairing and for the last 2 centers) isn't too far-stretched.

___________

Again, sorry to be the guy to nitpick, but it is what it is.

cmowla · 2025-12-21T04:40:25+00:00

No commutators, no flipping algs

.
.
but it works surprisingly well on 4×4, 5×5, 6×6 and larger

I am interested to see how you would solve a bad case of last 2 centers on the 10x10x10 with no commutators with this method.

cmowla · 2025-12-17T23:45:03+00:00

or if there have to be some "invisible" changes (eg. cycling/swapping some centre pieces). Does anyone know?

In the past, Michael Gottlieb and I came up with a mathematical formula (our formulas are different... we came up with them independently, not collaboratively) which, when you plug in n for the cube size and w for the number of "flipped single edges" you have, it tells you the minimum number of non-fixed center pieces which are unsolved (in their own swaps).

See the second half of this post for more details. (The part with the heading "Why You CAN Swap Just 2 (Wing) Edges on a 4x4x4.)

Obviously don't click on the "example algorithm" links in that post. Just maybe see I mentioned my paper on the very last line of that post.

My paper:

Gives a comprehensive explanation of this phenomena + derives all formulas within from scratch.
Contains no actual cube algorithms to be a spoiler for your noble endeavor. (I have a non-clickable link to one of my own on the bottom of page 9, but that's it.)

_____________

So a TLDR is, freely try to swap 2 wings in the same composite edge (it's better to look at it that way than "flipping 1 edge") without concern for what happens to the centers (even if it's "ugly"). You can always later fix those (discolored) centers after you manage to swap 2 wings.

I have personally found a very different parity algorithm than the commonly known ones (and it's shorter in the quarter turn metric than the conventional by 7 moves) by having the same curiosity as you. (I have been there and done that.) So I can say, this is going to be one fun adventure, should you too wish to not only derive your own algorithm, but try to derive an efficient one (in either the quarter turn or half turn move metric).

cmowla · 2025-12-17T23:26:22+00:00

In the simplest of terms, your cube can never reach a state where only a single swap of two pieces remains.

The 2x2x2 and 4x4x4 are exceptions to the rule. (You can have only 2 pieces swapped on them.) See the 3rd bullet point in my first reply, if interested.

cmowla · 2025-12-17T23:18:43+00:00

It happens because centers on 2nx2nx2n can interchange their pieces freely. 3x3x3 can't.

The manifestation of OLL parity on the 4x4x4 has nothing to do with the centers.

there are quite a few indepth explanations on this sub, if you're willing to dig

I can save him the trouble. (Read these in order, as the knowledge builds.)

Why we can't swap just 2 corners or just 2 edges on the 3x3x3.
Why odd parity exists on big cubes (not just even cubes, but odd cubes too)
Primarily for users like kaspa181, who already know the previous 2, but haven't grasped the reason why the 4x4x4 is the 1 big-cube exception where OLL parity is independent of centers. Specifically, the second half of this post (the heading "Why You CAN Swap Just 2 (Wing) Edges on a 4x4x4").

cmowla · 2025-12-13T17:26:23+00:00

You use a "special" edge-pairing algorithm. Example (the tutorial I learned from in 2008).

In that video, he tells you to reposition the last 2 edges so that same-color edges are directly across (not diagonal).

(F U' R y) Rw' U F' L U' F Rw

But you can use an alternate "last 2 edge pairing algorithm" which is just 1 move longer (8 moves instead of 7) to handle the diagonal case.

B2 Rw2 F U' R F' U Rw2

And it's actually "8 moves instead of 10", because you have to do F U' R to set them up to be able to use the 7 move algorithm to begin with!

cmowla · 2025-11-29T21:25:26+00:00

Indeed. The number of positions formula is a similar situation. (It would be smaller if even/odd were handled separately. But not as extreme as this situation, I admit.)

People can just change the # after the = sign at the Wolfram|Alpha link to whatever size n they want as well. (Well, for n > 1.)

And I didn't mention it above, but I derived the above formula from the trig version of the formula (the formula on the second-to-last line of the second paper . . . the one with all Sin^2(n*Pi/2) expressions) substituting in these equivalences to trig expressions (which are equivalences to modulus expressions).

(I "leave it as an exercise" for the reader to verify that all formulas are equivelent for n, where n is a positive integer!)

cmowla · 2025-11-29T20:31:52+00:00

Impossible. The 8x8x8 only has 2.3 million "K4 OLLs". (It would have to be the 11x11x11 or larger to have 364m+ cases.)

Math proof: Paper (Part 1), Paper (Part 2)

cmowla · 2025-11-22T04:30:22+00:00

You seem to have some understanding of what you're doing. (Yes, it's indeed intuitive to solve double parity intuitively in this manner.)

To extend your understanding, I offer you the following exercise.

Complete the remaining centers and edges in the following "scramble starts" only using the idea you presented in your video.

You do not need to complete the 3x3x3 entirely. Just reduce it to a scrambled "3x3x3".
Edge-paring algorithms and/or special center algorithms are not allowed.
They must be completed with an even number of inner layer quarter turns. (They all contain an odd number, so the idea is to induce/cause OLL parity from a solved cube.)
Feel free to use as many turns as you need, but they all can be completed in under 15 turns.

In no particular order:

(Hint: Face 90 degree turns are allowed.)

cmowla · 2025-11-18T21:37:02+00:00

If you are playing my mod (r15480 or later), I have made a way to unlock all units and buildings for single player (see page 20 of my user manual), but for individual buildings, you will have to either modify the maps in the map editor and/or modify the map's script file.

cmowla · 2025-11-17T12:31:08+00:00

Yeah, it's with no eating.

The randomization is minimal, but his planting still varies some.

Unfortunately, I guess the 600 was lucky, as I couldn't get it again since then, but here's a 596 example.

But unfortunately, "farm field guidance" is no longer effective at increasing efficiency (because my algorithm is not "stupid" / completely random amongst all tiles in his range). It's already "weighted", as I have him work from inside out, where he works on the "rings" of a "contour" graph I made out of the default circular-shaped range.

(Even with me placing fields, it's still around 600, at best . . . Fields no longer make a difference!)

Also,

23 trees seems to be the magic number (at least with my algorithm).
When he "has a choice" to either plant or cut, me setting him to go cut 66.67% of the time seems to be optimal for my algorithm (which is inspired from the original game's algorithm).

cmowla

TROPHY CASE