# 4x4 rU combinations



## Alex Shih (Jul 26, 2021)

Hi guys,

I'm currently working on a 4x4 method that relies a lot on r and U moves (r is right internal slice), and I need some help calculating how many positions are possible for certain steps. If a 4x4 is scrambled with just r and U2 moves and then the unsolved centers are solved just using r and U2 moves, how many positions are possible for the edge pieces? I know that the upper bound for this is 6 factorial since there are 6 unsolved edges, but I'm not sure how to take into account AUF (in this case I guess it would be AU2F). Is it just 6!/2=360?


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## Christopher Mowla (Jul 26, 2021)

To get you started,

God's number is 28. Also see @Ben Whitmore website page on it.

As Ben said, there are 274337280000 total configurations (on the regular 6 colored 4x4x4). There are 10 wing edges, 10 X centers, and 4 corners in reach of <U,r>. 10!*10!*4. But there are four X center color sets. Three colors have 2 X centers and one has 4. So you need to divide by 2!*2!*2!*4!.



Alex Shih said:


> how many positions are possible for the edge pieces?


You can have any of the 10! permutations of wing edges independently of any other piece (because you can do a 2-cycle: see here on the 4x4x4 parity algorithms wiki page and perhaps my video on how to make them by hand). In fact, God's number for the 8 wing edges in the last layer is 27. (See the entire thread.)

There are 4 different positions for the corners. Solved, U', U, U2. (AUF). You can (in theory) have all 4 positions with the corners permuted with nothing else (visually) unsolved, but I don't have any (short) algs for that. Repetition can be used to find long algorithms to do that, however.

The 10 X center pieces can be anywhere independently of the corners and wings (if we visualize swapping same color X centers to make the permutation legal regarding the parity dependence with the corners . . . if we take this statement literally, i.e., the 4x4x4 supercube).

We cannot just do a 2-cycle of X centers on the 4x4x4 supercube. But we can do a 2-cycle of X-centers if the corners are offset by 90 degrees. I don't have a special alg for it, but take [2R' U' 2R' U': (U 2R' U' 2R2)4] U' for example. As said above, you can (in theory) do any permutation of wing edges independently of any of the X centers and corners (for the non-supercube), so just imagine solving those wing edges back (and observe the algorithm on a 4x4x4 supercube to see that it switches the top back X centers).

I never found an algorithm to do just a 3-cycle of X centers in <U,r>, but the optimal solution is probably long.

Perhaps if you post images of specific types of subsets, we can give you a few examples on the number of positions for each, but clearly the number of positions reachable with <U, r> is much larger than what you originally anticipated.


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## xyzzy (Jul 26, 2021)

Alex Shih said:


> If a 4x4 is scrambled with just r and U2 moves and then the unsolved centers are solved just using r and U2 moves, how many positions are possible for the edge pieces? I know that the upper bound for this is 6 factorial since there are 6 unsolved edges, but I'm not sure how to take into account AUF (in this case I guess it would be AU2F). Is it just 6!/2=360?


I'm not sure what you mean by "AUF" here; are you ignoring whether the UL and UR bars are swapped? If yes, then AUF would cut the cases down by about half (not exactly half, because there are symmetrical states). If no, i.e. you care about solving UL and UR as well, then you shouldn't be reducing by AUF.

As for the actual number of permutations (without reducing by AUF), there is one nonobvious constraint that causes the number of permutations to only be 6!/6 = 120 rather than 6! = 720.

Imagine drawing lines between the UFl-UBl edges, the UFr-DBr edges, and the UBr-DFr edges. If you do only 2R and U2 moves, these three lines are locked to being in one of five configurations. In contrast, if you could do arbitrary permutations on those six edges, the three lines would have fifteen legal configurations instead. This argument shows that we _can't_ do arbitrary permutations, although it doesn't pin down exactly which ones we can do and which ones we can't. Showing that the exact number is 120 is a bit more complicated, to say the least.

---

As Christopher mentions above, if you _don't_ restrict to 2R and U2 moves in the scramble, then you can in fact get arbitrary permutations on all of the edge pieces. If you have reduction to ⟨2R, U2⟩ as one of the steps in your method, then you need to either solve this additional edge constraint during the reduction, or otherwise resolve it somehow (akin to solving parity at the end of reduction solves).


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## Christopher Mowla (Jul 26, 2021)

I also forgot to mention it, but also take a look at this thread (just for seeing some more algorithms). Specifically in this post, @cuBerBruce lists move optimal <U, r > algorithms for different 3-cycle of wing cases in the last layer.

I am not sure why your method uses algorithms strictly in < U, r >, but if you can afford to adjust it to allow R turns, algorithms can become shorter (and do a lot more things). Allowing u turns (so doing Uw on occasion when needed) allows it to have even more pieces in reach (and shorter algorithms).

*Almost always, move optimal algorithms in <U, r> are found with KSolve or KSolve++*. Algorithms in related restrictions (like allowing R turns) can also be found with that software. Example.


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## rokicki (Jul 26, 2021)

Christopher Mowla said:


> I never found an algorithm to do just a 3-cycle of X centers in <U,r>, but the optimal solution is probably long.


The optimal solution is indeed pretty long. Twenty moves!

U 2R' U' 2R2 U' 2R2 U2 2R2 U' 2R' U' 2R' U 2R2 U 2R2 U2 2R2 U 2R'


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