Cube Theory Quiz #1

[Stefan]

Stefan pointed out that I have a flawed definition of a 5-generator algorithm.

More like a flaw in what I suspect you wanted to ask (that you wanted a quarter turn). I probably wouldn't have called it "5 generator algorithm" in the first place, as that term to me simply sounds like an algorithm built using let's say {R,L,F,B,U}, not that it has to have a particular effect (like simulating a turn of the sixth side). Though maybe there's good reason to call it that and I just don't see it.

 

[Rune]

Is it correct or not correct, what Singmaster writes on p. 18 in his "Notes on..."? That Penrose has shown...
(The solution can be found on p. 27)

 

[cmhardw]

More like a flaw in what I suspect you wanted to ask (that you wanted a quarter turn). I probably wouldn't have called it "5 generator algorithm" in the first place, as that term to me simply sounds like an algorithm built using let's say {R,L,F,B,U}, not that it has to have a particular effect (like simulating a turn of the sixth side). Though maybe there's good reason to call it that and I just don't see it.

Focusing on vocabulary we can say that the entire cube is 5-gen (it's actually 2-gen according to Lucas, but we can say it's also 5-gen). I don't know what would be a good name for an algorithm that uses only 5 sides to simulate a quarter turn of the 6th side. An algorithm with this effect proves that the cube is 5-gen, but what would you suggest to call algorithms that have this property?

 

[porkynator]

I answered without reading any comment or searching for the answer.
8) and 9) were fun

Spoiler

1) Considering only the permutation of pieces, every state is reachable: using R U2 R' U' R U2 L' U R' U' L you can swap 2 corners; repeating this moves in different positions of the cube can lead to every possible state.
Considering orientation, only 1/3 of the states is reachable: consider only the top and bottom colors. Give each corner a value of 0 if the top/bottom color faces top or bottom (every corner has either a top or a bottom side, and every corners faces either top or bottom with one side), 1 if it is a clockwise turn away from being 0, 2 if it is a counterclockwise turn away from being 0. Assume twisting a corner on itself clockwise (counterclockwise) gives it a +2 (+1).
Now: <U, D, R2, L2, F2, B2> moves don't change the orientation of the corners; each <L, R, B, F> move gives +1 to 2 corners and +2 to 2 other corners. It's easy to see that the sum of the values of all corners must always be a multiple of 3, thus only those cases are reachable.

2) WLOG, let's assume the DLB corner is correctly placed and oriented (it can be placed and oriented using only <x, y, z>).
As stated in 1), the other 7 pieces can be permuted freely, giving 7! possibilities, while only 1/3 of the 3^7 different orientations are possible, giving 7!*3^6 = 3674160 possible states.

3) I assume the laws of the cube are known.
We can't assume one piece is already placed correctly as in 2), because we have fixed centers.
For edges: 2^12 possible orientations, but only half of the cases are possible, so 2^11; 12! possible permutations.
For corners: 3^8 possible orientations, but only 1/3 of the cases are possible, so 3^7; 8! possible permutations.
Only even permutations are possible, so we divide the result by 2 and get: 12!*8!*2^10*3^7 = 43252003274489856000.

4&5) I don't have a pyraminx and never thought about it

6) (Partial answer because I don't know the difference between OP and PP)
As in 3x3 we have corners/edges permutation parity (it can be proven similarly as I proved the corners orientation "parity" in 1)), in 4x4 we can have wings/centers parity (this also can be proven). Since each permutation of the center pieces in their face is undistinguishable if you are not solving a supercube, we can have OLL and PLL parity.

7) It's an A-9 commutator:
R2 as a setup move
R F R' as insertion
B2 interchange
R F' R' Inverse-insertion
B2 inverse interchange
R2 inverse steup
written with cancellations: R' F R' B2 R F' R' B2 R2
I can give a more detailed explanation on how commutator works, but I think it's useless now

8) Scramble: D
Solution:
R L' F' R' L //Corners
U' R' L F R L' //First edge
U R L' B2 R' L //Second edge
U F B' R2 F' B //Third edge
U' F' B L' B' F //Fourth edge
//LL skip

9) I assumed using well-know algs (as double T-perm for one 180° turn) was considered cheating

a) I'll call this A and use it also for b) and c): (M' U' M U)^5
b) A^2
c) I'll call this C:
(R2 U)^5
F' B U2
F' B D' F2 B2 U' F B' R2 F2 B2 L2 F B' U'
z2 y A
d) C z2 A

EDIT: Reading other people answers, I feel proud for having found an original solution for 8)
Also, I hope Chris is ok with me translating his post and posting it here (with reference to the original post, off course).

 

[Stefan]

Is it correct or not correct, what Singmaster writes on p. 18 in his "Notes on..."? That Penrose has shown...
(The solution can be found on p. 27)

I'd say it's correct, though I'm wondering why you're asking.

An algorithm with this effect proves that the cube is 5-gen, but what would you suggest to call algorithms that have this property?

If you insist on giving it a name, maybe "5-gen-prover"

 

[Rune]

I thought it had some connections with the Quiz here (8), but I was apparently mistaken.

 

[Stefan]

I thought it had some connections with the Quiz here (8), but I was apparently mistaken.

No, no, you're right, there is a connection, as it's pretty much exactly what Chris asked . That's why I'm confused, because I don't see why you would point to a question we already have. Was it because we were talking about how to write the question properly?

You're talking about Singmaster's problem 7D, right? "7D) The Five Generator Group: R. Penrose has shown that one generator can be ignored and we still get the whole group. I have been told that he has a 28 move process to produce the effect of one turn, using only the 5 other turns."

 

[Carrot]

another solution for 8:

Spoiler

L2 R2 U' L2 R2 U2 L2 R2 F2 L2 R2 B2 U F2 B2 L2 F2 B2 R2 U
F2 R2 L2 B2 L2 R2 U2 L2 R2 U L2 R2 U2 L2 R2 B2 L2 R2 F2
somewhat short one:
F B' R2 L2 F B' U F B' R2 L2 F B'

 

[cuBerBruce]

For those who have had fun with #8, I thought I would mention this similar question: Find a maneuver for F2 using only <U,D,L,R>.

I note that probably the most common edge orientation scheme is based upon the <U,D,L,R,F2,B2> subgroup. In this subgroup, the generators F2 and B2 are actually not needed. Just <U,D,L,R> generates the same subgroup. In fact, I have created a Hamiltonian circuit for this subgroup using only the moves { U,U',D,D',L,L',R,R' }. I used this as a stepping stone to come up with a Hamiltonian circuit for the whole cube group using only the moves { U,U',D,D',L,L',R,R',F,F' }.

The fact the F2 and B2 are not needed should be apparent to ZZ solvers. To come up with a sequence to do F2 without any F or B turns in a "brute force" way, take a solved cube, apply F2, then solve the cube ZZ style without any F/B turns. (Finish the line of the EOLine, such as with U' D R2 D', then finish the solve with <R,U,L>.)

 

[elrog]

^
I tried coming up with an algorithm for F2 using only 90 degree turns from the subgroup { U,U',D,D',L,L',R,R' } and doing it without using more than 2 D moves and 2 L moves. I found that it is impossible due to corner permutation and you have to either preform at least 4 twists on either D or L or do a 180 degree turn somewhere.

I thought experimenting with this was pretty interesting. Thanks for the challenge .

I have not proven this with a computer (partly because I don't know how) so feel free to prove me wrong.

 

[porkynator]

I just went for the F2, without caring about efficiency but without using pure ZZ (anyway, I use F2 for ZZ sometimes, and even F' R U R' U' R' F R sometimes for 2H).

Spoiler

Scramble: F2
R L' U2 L R' //All but 3 edges
U R2 U R2 L2 D R2 D' R2 L2 U R2 U2 R2 U' //Last 3 edges, can be seen as [U R2 U:[M2, U R2 U']]

Nice challenge

 

[Rune]

I know too well that I shouldn´t interfere in subjects that I don´t fully understand or that I don´t understand at all. But reading a certain post by Chris, I got the impression that it was not clear to him that we can get the whole group with the help of 5 generators, which had been shown by Penrose. That´s why I pointed (indirectly) on his "algorithm".

 

[Renslay]

For the F2 generation with <R,L,U,D> quiz, I have a solution with 22 HTM, only 2 quater turn moves on D side and 2 quater turns on L side, the rest is <U,R>:

Spoiler

Scramble: F2

U' D R2 U D' // solve corners
L' U2 R' // Setup for PLL
U2 R' U' R' U R U R U R' U // U-PLL on R side
R U2 L // Set down

 

[Stefan]

For those who have had fun with #8, I thought I would mention this similar question: Find a maneuver for F2 using only <U,D,L,R>.

Spoiler: Solution

Found a solution similar to Renslay (only did [sexy,lexy] cause I forgot that my normal U perms are <R,U>). Then found something curious with Cube Explorer:
(R' D R' U' R' U R2) (z y2) (R2 U R' U' R' D R') (z y2)
Note that the third part is the same as the first, just backwards.

 

[Ranzha]

Bump! I'm still eager to see your answers, Chris!