# Solving Rubik's cube in 36 quarter turns

In this paper we present a method of solving Rubik's cube in 36 quarter turns. The proof have many common facts with other methods presented on the forum. In able to prove our claim we have used the corner and edge permutation analysis done for the first time by Bruce Norskog.

We call the group of corner edge permutations of the cube for CEP and cube group for C. Let N be the normal subgroup that fixes cubies. Then we have a homomorphism hom:C->C/N=CEP such that hom(g)="permtutation of cubies done by g". Let a be the unique antipode in CEP. Every element in CEP can be written as x*a. Where x is an element requiring at most 18 quarter turns according to Norskog's analysis. There are 220 elements at distance 17 and 1 element at distance 18. So we could say that all elements of CEP except 220+1 elements can be written as x'*a where x' is an element requiring at most 16 quarter turns.

We will show that all elements g in the cube such that hom(g)=x'*a where x' is some element in CEP requiring 16 quarter turns can be solved in 36q or less. In order to do that we have computed solutions for all elements in the cube group having the same permutation of cubies
as following elements of CEP:

-U2*a (All but 14 elements solvable in 22q, the other 14 are local maxima 24q long)

-UD*a (All solvable in 22q)

-UD-1*a (All solvable in 22q)

-UL*a (All solvable in 22q)

-UL-1*a (All solvable in 22q)

Conjugating those elements with symmetries will yield solutions of all elements of CEP of the form x1*x2*a where x1,x2 is in the set X={U,L,F,B,R,D,U-1,L-1,F-1,B-1,R-1,D-1}. Let T be the set {x1*x2*a|x1,x2 in hom(X)}. One verifies that a is a symmetry i.e. a is in the group hom(M). So T is in fact equal to {a*x1*x2|x1,x2 in hom(X)} that is because a-1*x1*x2*a=x1'*x2'. Here x1,x2,x1',x2' are in hom(X). Every element g' of CEP except 220+1 can be written as g'=t*y where t is in T and y is at most 14 quarters long.
Now let g be an arbitrary element in the cube. Then we can find a h in the cube 14 quarter long in the cube such that hom(g*h) is in T. Every element u in the cube with hom(u) in T can be solved in 22q or 24q so we get following possibilities:

-g*h*f=id where f is 22q long and hom(f) is in T. Then g=f-1*h-1 has maximal length 14+22q.

-g*h*f=id where f is 24q long and hom(f) is in T. Then
g=f-1*h-1. Let h-1=x1*el. Where x1 is a generator and el an element of length max 13. Because f-1 is a local maxima it can be writte as d*x1-1 where d has length 23. Then g=f-1*h-1=d*x1-1*x1*el. Since d is of length 23 and el of length 13 we have that g is max 36q long.

Let us consider the remaining elements g in the cube with hom(g)=y'*a where y' is 17q or 18q. Bruce Norskog has kindly sent me the unique M elements in CEP that are 17q long:

U U D F' L D' R F' D' F' R' B' U' D' B' R F
U U D B R L D R' D' F' R B L' U F F L
U U D B U' B R L D' B' D B' R D L U' R'
U U D F R L B' L' F R L' B L' D F R F
U U D F U F' R' D R' F U' F R' B' U R' U'
U U D D B' L F' R D R D F F R' U D' L
U U D F U' L' F' U B R D R B L B U' F'
U U D F F D F L U L D' R U R F L F'
U U D D F U D F B R L B R R L L B
U U D D F U D F' B' R L F U U D D F
U U D D F R U' F' U' F D' F' B' U' L F F

I have personally checked that they are at distance 17 in CEP. To show that those elements g in the cube with hom(g)=y'*a(y' 17q long) are 36q long max we have computed solutions of all elements having following corner edge permutations:

-The identity(All but 25 elements solvable in 22q, the other 25 are local maxima 24q long)

-R B U L U F R D L D L' U B U U B' D' R F' U' L B (All solvable in 22q)

Call the last element b. Let S be the set of unique hom(M) elements above(distance 17 elements) and the antipode. One can verify that all elements in Sa={s*a| s in S} can be written either as b*z or as id*z where z is 14 quarter long max. So if we have an element g in the cube such that hom(g)=y'*a(y' 17q or 18q) then there exists a h in the cube such hom(g*h)=b or id and h 13q long. We get following cases:

-hom(g*h)=b. Then g*h*f=id and f max 22q and g=f-1*h-1 is 22+14q long max.
-hom(g*h)=id. Then g*h*f=id and f max 22q and g=f-1*h-1 is 22+14q long max.
-hom(g*h)=id. Then g*h*f=id and f is 24q long and g=f-1*h-1. Then we write h-1 as x1*el where el is some element 13q long max and x1 is a generator. Because f-1 is a local maxima it can be written as d*x1 with d 23q long and thus g=d*x1*x1-1*el=d*el which is 23+13q long max.

The proof is finished. Now some information about how the solution files are stored. Let m be in M(symmetry group). We have not stored solutions for two elemenents x and y having same corner edge permutation (i.e. having hom(x)=hom(y) ) with hom(m-1*x*m)=hom(y) or hom(m-1*x-1*m)=hom(y). The solutions can be found at my home page http://users.student.lth.se/f01sr/ .

I want to thank my advisor Gert Almkvist for much inspiration and help. My second advisor Victor Ufnarovski who patiently helped me correct a lot of errors. My family for supporting me. Of course i thank Bruce Norskog for the very hard work he have done on the corner edge permutation analysis. I thank Michael Reid for his solver that i used to show that the 24q positions are local maxima. I thank my cousin in law Daniel Grad who helped with some own made programs. Further i thank the people who contributed to the development of GAP and the people at Lunarc at Lunds University of Sweden for their computers that i used and especially to Per-Anders Wernberg for helping me start quickly.

## Comment viewing options

### This (and the below 28f) resu

This (and the below 28f) result seem pretty amazing to me; I'm shocked no one more knowledgeable than I am is commenting on them! If these results are correct (and I have no reason to believe they aren't), that means the diameter limits are now faceturn 20..28 and quarterturn 26..36, which is much tighter than the best results just a couple of months ago, correct? Jerry, Mark, what do you think?

### I think we are basically in a

I think we are basically in a new golden age of cube research, making an amazing amount of progress. I don't know if it's because we have a number of new people working on the problem as compared to Cube-Lovers days, or because we are a couple of generations further along with computer hardware. Maybe it's a little bit of both. But the progress is quite amazing.

### The 28f result

Oh I'm reading all the posts and yes I think it's a big step forward, definitely congratulations are in order. Correct me if I'm wrong but I don't think we have any position proven to be beyond 20 face turns so I don't think a lower bound of 28 face turns to be _too_ surprising. I'm no expert on lower bounds for the 3x3x3 cube. Lately I've been very busy making physical puzzles otherwise I'd be hacking away on some other subgroups. There's still a lot of unmined territory there... when I have some time I'll like to hack up God's Algorithm for the 3x2x2 which will be very easy. Maybe 4x3x3 is also doable?

### > God's Algorithm for the 3x2

> God's Algorithm for the 3x2x2

Well, I found the diameter to be 14f in 2002, but I would guess that "many" others had done it before! Assuming no errors, there were 576 positions antipodal to START.

Mike

### That agrees with the run I di

That agrees with the run I did this morning:
```STM    FTM-->
v     0 1  2   3   4    5    6     7     8     9    10    11    12    13  14  Total
0: 1 0  0   0   0    0    0     0     0     0     0     0     0     0   0      1
1: 0 6  0   0   0    0    0     0     0     0     0     0     0     0   0      6
2: 0 2 22   0   0    0    0     0     0     0     0     0     0     0   0     24
3: 0 0 12  81   0    0    0     0     0     0     0     0     0     0   0     93
4: 0 0  1  64 276    0    0     0     0     0     0     0     0     0   0    341
5: 0 0  0  12 298  820    0     0     0     0     0     0     0     0   0   1130
6: 0 0  0   0  96 1140 2100     0     0     0     0     0     0     0   0   3336
7: 0 0  0   0   8  517 3624  4600     0     0     0     0     0     0   0   8749
8: 0 0  0   0   0   50 1652  8964  8156     0     0     0     0     0   0  18822
9: 0 0  0   0   0    0   66  3456 17056 11984     0     0     0     0   0  32562
10: 0 0  0   0   0    0    0    68  6296 23956 14512     0     0     0   0  44832
11: 0 0  0   0   0    0    0     0    60  8456 25528 15232     0     0   0  49276
12: 0 0  0   0   0    0    0     0     0   308  6480 27648  9792     0   0  44228
13: 0 0  0   0   0    0    0     0     0     0   696  6208 16824  4624   0  28352
14: 0 0  0   0   0    0    0     0     0     0     0   704  2096  6000 576   9376
15: 0 0  0   0   0    0    0     0     0     0     0     0   312   480   0    792
Total: 1 8 35 157 678 2527 7442 17088 31568 44704 47216 49792 29024 11104 576 241920
```
Jaap's Puzzle Page: http://www.geocities.com/jaapsch/puzzles/

### Great. My own program was ju

Great. My own program was just a quick application to try out different methods of indexing permutations, and wasn't checked carefully. Hadn't considered the other metric, either.

### > Correct me if I'm wrong but

> Correct me if I'm wrong but I don't think we have any position
> proven to be beyond 20 face turns so I don't think a lower bound
> of 28 face turns to be _too_ surprising.

In my opinion, the actual group diameter is likely to be close to 20, so the fact that it is no more than 28 is indeed not surprising. The fact that it has been _proven_ to be so, is.
I am also very happy to see that the bounds for qtm are now much better. Calculations on that side seemed to have fallen behind, but now they are pretty closely matched:

> faceturn 20..28
> quarterturn 26..36

Note that the ratios are nearly the same:
20:28 ~ 26:36
or equivalently
20:26 ~ 28:36
20*36=720, 26*28=728

If anything, it indicates to me that the htm bounds will probably be improved next. Especially because the 26 qtm lower bound is such an outlier.

> when I have some time I'll like to hack up God's Algorithm for
> the 3x2x2 which will be very easy.

I can do that easily too with my current generic solver. I'll post the result here later.

> Maybe 4x3x3 is also doable?

Not likely. It is like 2 domino puzzles stacked together - the outer layers form one domino, the inner layers the other. While these can be solved independently, much like layers of a Masterball, they share moves so that they can be solved simultaneously in fewer moves. A God's Alg calculation must therefore treat the puzzle as a whole.
Anyway, the number of positions is domino^2 = 8!^4. This can of course be reduced a little by symmetry, but it is still out of reach for a complete calculation.

Jaap's Puzzle Page:
http://www.geocities.com/jaapsch/puzzles/