This is Part 4 of a 4part blog post on the mathematics of the 4x4 Rubik's Cube, its relation to algorithms, and some curious properties of Rubik's Cubes.
See Part 1 of this blog post here: Part 1: Representations
See Part 2 of this blog post here: Part 2: Permutations
See Part 3 of this blog post here: Part 3: Factoring Permutations
You are currently reading Part 4 of this blog post: Part 4: Sequence Order
Table of Contents
Introduction
Order of a Sequence
As a reminder of our overarching goal: starting with a 4x4 Rubik's Revenge cube, an arbitrary sequence of moves will scramble the faces of the cube; but if that move sequence is repeatedly applied, eventually the cube will return to its solved state.
The simplest example is rotating a single face: after applying the rotation move four times to any face of a solved cube, the cube will return back to the solved state.
This is also true of more complicated move sequences, such as
U R U' R'
, which returns the cube back to its original state
after 6 applications, or the move sequence U R
, which
must be applied 105 times before the cube returns back to
its original solved state.
Our goal is to predict this number: given a move sequence, how many times must that move sequence be applied to a solved cube to return the cube back to its solved state?
This number is called the order of a sequence.
What We Have Covered So Far
In prior posts, we have covered a number of key topics that this post will synthesize.
We started Part 1 by discussing ways of representing the Rubik's Revenge cube, and we settled on a 96tuple representation indicating which faces had moved to what locations.
That led us to Part 2, in which we discussed the tworow notation for the 96tuple representing the cube, and demonstrated the utility of this representation by showing how moves and move sequences would lead to permutations that could be written as 96tuples using the tworow notation.
In Part 3, we covered some key theoretical results following Donald Knuth's Art of Computer Programming which allowed us to develop a permutation algebra to describe the effects moves have on the cube. We concluded the previous post with an algorithm for factoring permutations into their intercalation products, and hinted that these permutation factors were central
Factoring Rubik's Cube Permutations
Factoring Permutations: A Review
In Part 3 of this series of blog posts, we looked at an example multiset permutation of characters. Here it is written using the tworow notation:
We covered a technique for factoring this permutation into independent cycles of faces,
and shared Python code to perform this operation. The resulting factored permutation was:
Factoring Rubik's Cube Permutations
To factor a Rubik's Cube permutation, we apply Algorithm A from the prior post to the tworow 96tuple representation of the Rubik's Cube after it has had the move sequence applied once.
(Note that we only need to apply the sequence to the cube once, even if the order of that sequence is in the tens of thousands.)
Let's look at a few move sequences for some examples:
Computing the Order of Sequence R
We begin with the solved state, and apply the move R to the cube. The result is the twoline representation:
(01 02 03 04 05 06 07 08 09 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96)
(01 02 03 36 05 06 07 40 09 10 11 44 13 14 15 48 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 84 37 38 39 88 41 42 43 92 45 46 47 96 61 57 53 49 62 58 54 50 63 59 55 51 64 60 56 52 16 66 67 68 12 70 71 72 08 74 75 76 04 78 79 80 81 82 83 77 85 86 87 73 89 90 91 69 93 94 95 65)
Now, we can carry out the Algorithm A procedure on this tworow representation. When we do that, we will find that there are a large number of oneelement independent factors; these are the faces that do not move during the move sequence R.
Here is a list of factors that are found by Algorithm A:
Factor sizes: {1, 4}
Factors:
[36, 84, 77, 4]
[40, 88, 73, 8]
[44, 92, 69, 12]
[48, 96, 65, 16]
[61, 64, 52, 49]
[57, 63, 56, 50]
[53, 62, 60, 51]
[58, 59, 55, 54]
Independent Faces: [1, 2, 3, 5, 6, 7, 9, 10, 11, 13, 14, 15, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 37, 38, 39, 41, 42, 43, 45, 46, 47, 66, 67, 68, 70, 71, 72, 74, 75, 76, 78, 79, 80, 81, 82, 83, 85, 86, 87, 89, 90, 91, 93, 94, 95]
Least common multiple: 4
The largest set of faces that are exchanged is 4, and the smallest is 1. No other groups of faces being exchanged have any other sizes. This means that if we apply the sequence 4 times, each of those groups of faces being interchanged will have returned to their original state.
This tells us what we already knew: that if we apply the sequence "R", it rotates groups of pieces in a sequence of 4 moves each, so overall the order of this permutation is 4  if we apply the sequence R to a solved 4x4 Rubik's Revenge cube 4 times, the cube will return to the solved state.
To formalize this, if we have cycles with arbitrary lengths, we must apply the sequence a number of times equal to the least common multiple of each factor's size. (For example, if we had a cycle of length 3 above, the cycle order would have been 12  because the sequence must be applied 12 times before the 4cycle face exchanges "sync up" with the 3cycle face exchanges.)
Let's look at a slightly more complicated move sequence to illustrate this point.
Computing the Order of Sequence U R U' R'
As before, we begin by applying the move sequence once to a solved cube to generate the tworow ntuple representation:
(01 02 03 04 05 06 07 08 09 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96)
(01 02 03 77 05 06 07 73 09 10 11 69 16 12 08 20 17 18 19 36 21 22 23 24 25 26 27 28 29 30 31 32 49 50 51 33 37 38 39 40 41 42 43 44 45 46 47 48 13 56 60 64 53 54 55 34 57 58 59 35 61 62 63 04 96 66 67 68 14 70 71 72 15 74 75 76 65 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 52)
Next, we factor this permutation using Algorithm A:
Factor sizes: {1, 3, 6}
Factors:
[77, 65, 96, 52, 64, 4]
[73, 15, 8]
[69, 14, 12]
[16, 20, 36, 33, 49, 13]
[50, 56, 34]
[51, 60, 35]
Independent Faces: [1, 2, 3, 5, 6, 7, 9, 10, 11, 17, 18, 19, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 53, 54, 55, 57, 58, 59, 61, 62, 63, 66, 67, 68, 70, 71, 72, 74, 75, 76, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95]
Least common multiple: 6
This time, we get a couple of cycles with different lengths. We have four cycles of length 3, and two cycles of length 6, plus many cycles of length 1 (the unpermuted faces).
The LCM of 3 and 6 is 6, so the overall order of the
move sequence U R U' R'
is 6.
Computing the Order of Sequence U R
The last sequence we'll look at is the move sequence UR.
This particular permutation represents an interesting corner case: in Part 1 of this post, when we came up with our tuple representation for the cube, we treated each face as being noninterchangeable, by giving each face a unique number. This means that, for example, we cannot swap two arbitrary red faces, since they are attached to other faces via a double edge or a corner piece.
This assumption does not hold for faces in the center of the cube. Because center faces are not attached to any other faces (mechanically speaking), the four distinct integers representing four colored faces can actually be interchanged.
This plays out with the sequence U R
as follows:
We start with the twoline representation of the ntuple:
(01 02 03 04 05 06 07 08 09 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96)
(13 09 05 01 14 10 06 02 15 11 07 03 48 44 40 36 33 34 35 84 21 22 23 24 25 26 27 28 29 30 31 32 61 57 53 49 37 38 39 88 41 42 43 92 45 46 47 96 16 66 67 68 62 58 54 50 63 59 55 51 64 60 56 52 17 18 19 20 12 70 71 72 08 74 75 76 04 78 79 80 81 82 83 77 85 86 87 73 89 90 91 69 93 94 95 65)
We can factor this tuple as follows:
Factor sizes: {1, 3, 4, 7, 15}
Factors:
[13, 48, 96, 65, 17, 33, 61, 64, 52, 68, 20, 84, 77, 4, 1]
[9, 15, 40, 88, 73, 8, 2]
[5, 14, 44, 92, 69, 12, 3]
[10, 11, 7, 6]
[36, 49, 16]
[34, 57, 63, 56, 50, 66, 18]
[35, 53, 62, 60, 51, 67, 19]
[58, 59, 55, 54]
Independent Faces: [21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 37, 38, 39, 41, 42, 43, 45, 46, 47, 70, 71, 72, 74, 75, 76, 78, 79, 80, 81, 82, 83, 85, 86, 87, 89, 90, 91, 93, 94, 95]
Least common multiple: 420
However, the adventurous cuber will find, when actually carrying out this move sequence, that the order is in fact 105, and not 420.
The reason the predicted cube order is 4 times larger than expected is because, after 105 applications of the move sequence, the cube has not actually returned to its original state, but the only remaining faces that are scrambled are center faces, which are in fact interchangeable.
Note this group of 4 faces that are permuted:
[10, 11, 7, 6]
These are the four center squares from the U
face.
If we exclude this group (treating 10, 11, 7, and 6 as
perfectly interchangeable), the length of all factors
no longer contains 4:
Factor sizes: {1, 3, 7, 15}
Including the 4, we had
LCM(1,3,4,7,15) = 420
but excluding the 4, we get:
LCM(1,3,7,15) = 105
Systematically, we can search for any groups that contain only faces from the center, and treat 1 such group of length n as n groups of length 1 (not contributing to the order of the move sequence).
This provides an interesting contrast between the 4x4 Rubik's Revenge cube, in which any center faces may be interchanged with any other center faces, and the 3x3 Rubik's Cube, in which the center faces always remain fixed in relation to one another.
Computing the Order of Sequence Uw Rw
We mentioned in Part 1
that the move notation Uw
or Dw
indicates a
quarter clockwise turn of two layers of a face,
not one. We can write the permutation that results
from the move sequence Uw Rw
as:
(01 02 03 04 05 06 07 08 09 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96)
(13 09 05 01 14 10 06 02 47 43 39 35 48 44 40 36 33 34 83 84 37 38 87 88 25 26 27 28 29 30 31 32 61 57 53 49 62 58 54 50 41 42 91 92 45 46 95 96 16 15 67 68 12 11 71 72 63 59 55 51 64 60 56 52 17 18 19 20 21 22 23 24 08 07 75 76 04 03 79 80 81 82 78 77 85 86 74 73 89 90 70 69 93 94 66 65)
Factoring this permutation, we get:
Factor sizes: {1, 3, 15}
Factors:
[13, 48, 96, 65, 17, 33, 61, 64, 52, 68, 20, 84, 77, 4, 1]
[9, 47, 95, 66, 18, 34, 57, 63, 56, 72, 24, 88, 73, 8, 2]
[5, 14, 44, 92, 69, 21, 37, 62, 60, 51, 67, 19, 83, 78, 3]
[10, 43, 91, 70, 22, 38, 58, 59, 55, 71, 23, 87, 74, 7, 6]
[39, 54, 11]
[35, 53, 12]
[40, 50, 15]
[36, 49, 16]
Independent Faces: [25, 26, 27, 28, 29, 30, 31, 32, 41, 42, 45, 46, 75, 76, 79, 80, 81, 82, 85, 86, 89, 90, 93, 94]
Least common multiple: 15
Several groups of 3 faces and of 15 faces, respectively, are permuted,
giving an LCM of 15. Thus, the order of move sequence Uw Rw
is 15.
We'll look at the factoring of one last sequence: U Rw
.
Computing the Order of Sequence U Rw
Here is the permutation representing the permutation resulting from the sequence U Rw:
(01 02 03 04 05 06 07 08 09 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96)
(13 09 05 01 14 10 06 02 47 43 39 35 48 44 40 36 33 34 83 84 21 22 23 24 25 26 27 28 29 30 31 32 61 57 53 49 37 38 87 88 41 42 91 92 45 46 95 96 16 15 67 68 62 58 54 50 63 59 55 51 64 60 56 52 17 18 19 20 12 11 71 72 08 07 75 76 04 03 79 80 81 82 78 77 85 86 74 73 89 90 70 69 93 94 66 65)
Factoring this permutation, we get:
Factor sizes: {1, 3, 4, 10, 15, 16}
Factors:
[13, 48, 96, 65, 17, 33, 61, 64, 52, 68, 20, 84, 77, 4, 1]
[9, 47, 95, 66, 18, 34, 57, 63, 56, 50, 15, 40, 88, 73, 8, 2]
[5, 14, 44, 92, 69, 12, 35, 53, 62, 60, 51, 67, 19, 83, 78, 3]
[10, 43, 91, 70, 11, 39, 87, 74, 7, 6]
[36, 49, 16]
[58, 59, 55, 54]
Independent Faces: [21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 37, 38, 41, 42, 45, 46, 71, 72, 75, 76, 79, 80, 81, 82, 85, 86, 89, 90, 93, 94]
Least common multiple: 240
The order of the move sequence U Rw
is 240.
Code
The code that forms the permutation tuple for a given move sequence and performs the factoring of that tuple is in sequence_order.py.
The sequence_order.py
file utilizes the
dwalton76/rubikscubeNxNxNsolver
library from Github to apply the move sequence once to a cube
to determine the resulting permutation tuple. It then factors
this tuple into products and finds the LCM of their lengths.
The code in sequence_order.py
is grouped into functions,
with the key funtion being factor_permutation(top, bottom)
, which
takes the top and bottom rows of the tworow representation
of a move sequence's permutation.
The method then performs the factoring procedure covered in Part 3.
Here is the body of the method:
def factor_permutation(perm_top,perm_bot):
"""
Factor a permutation into its lowest terms
"""
MAX = 96
# Need a way to also mark them as used... bit vector
used_vector = [0,]*len(perm_top)
i = 0
start = perm_top[0]
used_vector[0] = 1
factors = []
# If we still have values to pick out:
while(0 in used_vector):
factor = []
while(True):
used_vector[i] = 1
leader = perm_top[i]
follower = perm_bot[i]
i = perm_top.index(follower)
while(used_vector[i]==1):
i += 1
if(i>=MAX):
break
if(i>=MAX):
break
elif(follower==start):
break
else:
factor.append(follower)
# add start to end
factor.append(start)
factors.append(factor)
try:
#import pdb; pdb.set_trace()
i = used_vector.index(0)
start = perm_top[i]
except ValueError:
break
return factors
This was called by the method applying move sequences to the Rubik's Cube to obtain the tworow permutation corresponding to the move sequence of interest.
Project Conclusions
In addition to being interesting, this project led to some deep insights into the workings of the Rubik's Cube and ways to think about move sequences.
More than that, the Rubik's Cube is a toy that provides real insight into combinatorics and group theory. The concept of order, and the process of thinking through different representations of the cube and their consequences for the implemetation of the final algorithm, provide good practice for problems in other, related domains.
This project began with a simple question. While playing with the Rubik's Cube, we discovered this property of cycles (it is actually difficult to miss, even when learning the beginner method, as many of the move sequences involved in the beginner method have small orders, so it is easy to see them repeat.) The question we set out to answer was, given an arbitrary sequence, can we determine the order of that sequence?
The key to answering this question ultimately lies in the representation of the permutations; the right representation makes finding the order possible. but it took some trial and error with different representations before discovering the right approach.
To anyone who has played with the Rubik's Cube before, it seems natural that there would be some way to represent moves applied to the cube in some kind of algebraic terms. The intercalation product was the key concept for developing a permutation algebra. Knuth's Algorithm A was the key concept for factoring permutations into their respective independent cycles.
Once an algorithm to factor permutations was developed, the rest was a straightforward calculation of the LCM of the lengths of each factor.
The project was computationally challenging; recursion was required to implement Algorithm A, the Rubik's Cube solver had to be modified, and there were many bugs along the way.
The procedure we used here can be applied to other problems. Our procedure was:
 Find a proper, convenient representation for the system state
 Break down the variations of the system into simple cases or steps
 Move away from the specific system, and keep the approach mathematially general. This is by far the the most important step!
 Study the literature and solutions to problems, to become familiar with different ways of representing a problem. Different problems lend themselves well to different representations, so the more familiar you are with different representations, the more problems you'll be able to tackle.
 The only way to get familiar with different problemsolving approaches is through practice. It helps to start with easier problems, both because you can score some quick points and feel more confident, and also because combinatorics and group theory problems often tend to appear simple, but deceptively so. The devil is in the details.
References

"Rubik's Cube". Charlesreid1.com wiki, Charles Reid. Edited 25 January 2017. Accessed 25 January 2017. <https://charlesreid1.com/wiki/Rubiks_Cube>

"Rubik's Revenge". Charlesreid1.com wiki, Charles Reid. Edited 25 January 2017. Accessed 25 January 2017. <https://charlesreid1.com/wiki/Rubiks_Revenge>

"Rubik's Cube/Tuple". Charlesreid1.com wiki, Charles Reid. Edited 25 January 2017. Accessed 25 January 2017. <https://charlesreid1.com/wiki/Rubiks_Cube/Tuple>

"Rubik's Cube/Permutations". Charlesreid1.com wiki, Charles Reid. Edited 25 January 2017. Accessed 25 January 2017. <https://charlesreid1.com/wiki/Rubiks_Cube/Permutations>

"Github  dwalton76/rubikscubeNxNxNsolver". dwalton76, Github Repository, Github Inc. Accessed 11 January 2017. <https://github.com/dwalton76/rubikscubeNxNxNsolver>

"Rubik's Cube NxNxN Solver". Git repository, git.charlesreid1.com. Charles Reid. Updated 25 January 2017. <https://charlesreid1.com:3000/charlesreid1/rubikscubennnsolver>

"Rubiks Cube Cycles". Git repository, git.charlesreid1.com. Charles Reid. Updated 25 January 2017. <https://charlesreid1.com:3000/charlesreid1/rubikscubecycles>
Appendix
That concludes our discussion of computing the order of move sequences on a Rubik's Cube. There are many move sequences, and many orders, ranging from 1 or 2 up to nearly 100,000. We plan to assemble a web site to help readers explore some move sequences and their orders  so check back soon...