Here are three methods for manipulating the Up layer using conjugation. They're not optimal but they're easy to remember.
Three corner cycle
F D' F'
F D F'
F D' F'
F D F'
Three corner cycle with corner twist
F D' F'
R' D' R
R' D R
F D F'
Edge-corner three cycle
L' F L
I extended the above analysis to the Quarter Turn Metric and the Slice Turn Metric.
(I'm done trying to use the table editor here. You'll have to copy and paste into a spreadsheet if you're interested. The columns are the same as above.)
6 192 8 8
8 1,536 64 64
I knocked together some code today and enumerated the set of cube states with five unsolved edge cubies. As it turns out the above calculation is not valid. There are 462,528 cube states with 5 unsolved edges not 712,800. The latter number includes a lot of duplicates of states with less than...
You are correct. My calculation includes cases which are not 5 cycles. There are three cycles + two cubies flipped in place, etc. The requirement to preserve edge orientation is ambiguous since that depends on how one defines edge orientation.
I did some back of the envelope calculations:
The number of sets of 5 edge cubies:
12! / 7! = 95,040
Each of these edge sets may be permuted:
5! / 2 x 2^4 = 60 x 16 = 960
95040 x 960 = 91,238,400 edge permutations of five or fewer cubies.
12! / 8! x 4!/2 x 2^3 =...
Another definition of edge flip counts the number of quarter turns required to return the edge piece to the solved position and orientation. Even flip requires an even number of q-turns. Odd flip requires and odd number of q-turns.
I chose the above macros in the context of what is the minimum a person needs to know in order to solve the cube. You can tell a novice that all he has to learn is these five macros and he can solve the cube simply by repeatedly applying them.
I have considered the problem of solving the cube using a small set of primitive macros such as:
A tri-corner swap
F' R L F' R F L' F' R' F R' F
A tri-edge swap
F' U F' U' F' U' F' U F U F2
A tandem edge+corner swap
R U2 F' R' F U' F' R F U' R' U'
A corner twist
F R U F U' B U F' U' R'...
The positions a 3x3x3 may be placed in by rotating the faces may be represented as a mathematical group. The size of this group is 43,252,003,274,489,856,000 and may be calculated in the manner shown in the video. This is the "real" size of the group.
When one is performing "god's algorithm"...
If by Picture cubes you mean 3x3x3 cubes where the orientation of the center cubies matter, then the diameter is not known but it is at least 24f. Superflip requires 24 face turns: R L' U R U F R' L2 U F D' F2 R L' B2 D B' U' L R2 B' U' L' U'
In the 3 x 3 x 3 group modeled on the qtm metric or the ftm turn sequences do compose like group elements. The product of two group elements may be found by con-catenating generator turn sequences. The inverse is found by inverting the turn sequence. With the stm this is not the case since...
Ok, let's think this through.
There is a mathematical group, G, composed of all the arrangements the 4x4x4 puzzle pieces may be placed in by combinations of the puzzle moves. The group G has a subgroup, S, composed of all the even parity arrangements which may be produced by taking the solved...
Ok, I get you now. Good example. When you have identical pieces you're not dealing with a mathematical group but rather a coset space of a mathematical group. Inverses are not defined for coset spaces unless you're dealing with cosets of a normal subgroup. The 4x4x4 does have identical pieces...
I viewed the video and have some thoughts:
In order to model the 4x4x4 cube as a mathematical group one must define a preferred orientation. Usually one defines the preferred orientation as that which places the DLB cubie properly oriented in the DLB slot. The standard turn set for the 4x4x4...
I don't follow the point you are making here. I repeat that the states of the 4x4x4 can indeed be mapped to a permutation group and the inverse is well defined. Now if you take a non-optimal solution sequence for a randomly selected 4x4x4 cube state and apply it to a pristene cube you will not...
I'm not sure that that is true. The 4x4x4 states certainly may be mapped to a permutation group. In the 4x4x4 cube you have the complication that the turn set allows each cube state to be produced in any of 24 orientations. This is shared by the 2 x 2 x 2 cube, the void cube, and the 3 x 3...
Here is some code that I use to generate random 3x3x3 cube states. It is
written in Objective C which you may be familiar with if you're working in iOS.
Everything is fairly generic until the end where I translate the position and
orientation info into my working cube representation.
Not efficient but it is human readable. Also, it may be directly used to represent the cube as a facelet permutation. One may then use permutation multiplication to model cube manipulations.
Index.........12 34 56 78 90...