# Cube Theory Quiz #1



## cmhardw (Oct 17, 2013)

Hi everyone! I've been wanting to make a thread like this for a while now. There are a lot of neat cube theory results that are accessible to those who are interested in the cube, and who have a high school or advanced middle school math background. I've compiled a list of results that I find to be useful as well as ones that I think are very interesting. The level of questions here are for those who are just starting out with the cube who would like to learn more about the theory behind twisty puzzles. If you know how to solve the cube or are beginning to learn how to speedsolve, and you find yourself interested in the _why_ questions related to the cube, then you may enjoy this thread!

*How the quiz works*

The quiz consists of a series of questions loosely related to each other. The questions will vary in difficulty, as well as in effort level to answer the question. *Answer the questions you find to be interesting!*

When answering a question, please do so within spoiler tags. [noparse]


Spoiler



Your answers


[/noparse]

If you are stuck on a question, ask! Post a reply in this thread and talk it out with others who are also following this thread. We do enough competing on this forum in the forum competition area, so there will be no points awarded for the quiz. You may, however, receive Kudos from the theory experts on the forum for clever or well thought out answers to questions.

*In short:
Post your answers to these questions in spoiler tags. Ask questions without using spoiler tags. Answer others' questions without using spoiler tags.*

If people enjoy this thread, then I will come up with new questions.

*Cube Theory Quiz #1:*

*Question #1)* Read Ryan Heise's page on cube laws for the 3x3x3. Now, write out the cube laws for the 2x2x2.


*Question #2)* Calculate the number of legal positions to a 2x2x2. Give a brief explanation for each term in your calculation. Simply providing a numerical answer to this question does not sufficiently answer the question.


*Question #3)* Calculate the number of legal positions to a 3x3x3. Give a brief explanation for each term in your calculation. Simply providing a numerical answer to this question does not sufficiently answer the question.


*Question #4)* Write out the "Cube Laws" for the pyraminx. The "Cube Laws" are a list of any restrictions the puzzle must satisfy to be in a "legal" state.


*Question #5)* Calculate the number of legal positions to a pyraminx. Include the rotations of the outer tips in your calculation. As usual, give a brief explanation for each term in your calculation. Simply providing a numerical answer to this question does not sufficiently answer the question.


*Question #6)* When solving the 4x4x4 using the standard reduction method (i.e. solve center groups, then pair up edge groups, then solve the puzzle as a 3x3x3), what is the cause of OLL parity? (Hint: It does not just "happen", you created it. Tell me how.)


*Question #7)* Explain why the speedsolving A-perm alg *R' F R' B2 R F' R' B2 R2* works. You can answer this question in two ways. Either A) Explain how this algorithm changes the positions of the affected corners, then as a separate part explain how it preserves their orientation, or B) explain the structure of this algorithm and classify it for what type of algorithm it is.


*Question #8)* Come up with a 5 generator algorithm on 3x3x3 that is different from my signature and explain how it works. A 5-generator is defined as an algorithm that will use 5 sides of the cube to simulate a turn of the 6th side. Try the algorithm in my signature on a solved cube to get an idea of what this looks like.


*Question #9)* Come up with an algorithm to rotate two centers on a 3x3x3 supercube (if you don't have a supercube, make one!):
a) one 90 degrees clockwise, the other 90 degrees counter-clockwise
b) both 180 degrees
c) both 90 degrees clockwise
d) Come up with an algorithm to rotate exactly one center on a 3x3x3 supercube by 180 degrees.

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If there is enough interest in this thread, then I will post my answers to each of these questions by Thanksgiving weekend.

Happy cubing!


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## TDM (Oct 17, 2013)

cmhardw said:


> When answering a question, please do so within spoiler tags. [ s p o i l e r ] Your answers [ / s p o i l e r ]


You could use [noparse][noparse]


Spoiler



Your answers


[/noparse][/noparse].



Spoiler: My answers



*1.* Only one third of the corner orientations are reachable. (wasn't sure how much detail was needed for this question)
*2.* (8!*(3^7))/(6*4)=3,674,160
Explanation: 8! is the number of permutations. Imagine assembling a 2x2, putting pieces in in the same order (i.e. a piece in DBL, then DBR, then DFL, then DFR, then UBL, then URB, then UFL, then UFR) each time. There are 8 possible pieces to go in the first position, then 7 in the second place for each first piece etc. so the total number of permutations is 8x7x6x5x4x3x2x1, which is 8!. 3^7 is the number of possible orientations: for 7 of the corners, their orientation has no effect on the others, and as they have 3 possible orientations the number of possible different orientations is 3^7. The last corner's orientation is decided by the orientation of the others, so it isn't 3^8. Dividing by 6*4 is because of different orientations: 6 possible faces can be on U and for each one there are 4 different faces that can be on F.
*3.* ((8!*(3^7))*(12!*(2^11)))/2=43,252,003,274,489,856,000
Explanation: (8!*(3^7)) is the number of positions for corners, as explained above. As centres are fixed, we do not divide by the number of orientations.
(12!*(2^11)) is for the edges, using the same concept as for the corners of a 2x2. 12! is the number of permutations, and 2^11 is the number of orientations (again remembering that the last piece's orientation is decided by the others'). We now have to divide by 2, as only half the number of permutations are possible.
*4.* I only got one recently and don't use it very often, so I don't know much about pyraminx. But I think only half the edge permutations and edge orientations are possible.
*5.* Not more pyra  All calcs are done for centres being in the same position and assuming my answer to question 4 was correct (don't know if it is, but I hope so).
Tips: 3^4 (3 orientations for each of 4 tips)
Centres: 3^4 (3 orientations for each of 4 centres)
Egdges: (6!/2)*(2^5) (6! permutations, divide by two because last two edges' permutations are decided by the permutations of the other four, 2^5 orientations)
So (3^4)*(3^4)*((6!/2)*(2^5))=75,582,720
*6.* Is it by scrambling then solving centres+edge pairing using an odd number of turns of inner layers? I think I may have seen this before, but can't remember where and whether this was actually what was said or not.
*7.* R' F R' B2 R F' R' B2 R2 -> R' (F R' B2 R F' R' B2 R) R -> R' [F,R' B2 R] R -> *[R':[F,R' B2 R]]*
This A perm is a commutator with a conjugate (can you say it's a conjugated commutator? Idk much about them). It sets up the pieces to a position where they can be solved with a commutator with an R', then does the commutator [F,R' B2 R], then does an R to undo the setup. The commutator [F,R' B2 R] cycles UBL->FUR->FDR(->UBL). The F sets up UBL and FUR to be swapped with R' B2 R. F' then sets up FDR to be swapped with what is in UBL (which was originally in FUR). So the piece that started in UBL has now been put into FUR, what was in FUR is now in FDR and so what was in FDR is now in UBL.
*8.* F B L2 R2 B' F' U F B R2 L2 B' F'
Can also be written as [F B L2 R2 B' F':U]. The first part (F B L2 R2 B' F') moves all pieces on D to the U face so that they are correctly permuted when compared to each other. Turning U then simulates turning the pieces as if they were on the D face. F B R2 L2 B' F' is the inverse of the conjugate: this returns the pieces to the D layer in their new permutation and resolves the other pieces.
*9.* One CW, one CCW: (R2 U R U R' U' R' U' R' U R')3 turns the U centre CCW and the R centre CW
Both 180: [(L' U L U)5:z'] turns R and U 180 degrees. It's the alg for turning one centre 180 degress done on both sides.
Both CW: (R2 U R U R' U' R' U' R' U R')3 (L' U L U)5 is a combination of the first and last parts of this question.
One 180 degrees: (L' U L U)5 turns the U centre 180 degrees


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## cmhardw (Oct 17, 2013)

TDM said:


> You could use [noparse][noparse]
> 
> 
> Spoiler
> ...



Thanks TDM! I didn't know about that, I edited my first post to include this.


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## AndersB (Oct 17, 2013)

My entry for the ones I could do now. 


Spoiler: Answers



1. All permutations are reachable. Only one third of the corner orientations are reachable. Combining these laws, only 1/3 of the conceivable cube states are reachable by legal moves.
4. Only half of the permutations are reachable. Only half of the edge orientations are reachable. Combining these laws, only 1/4 (1/2 * 1/2) of the conceivable cube states are reachable by legal moves.
6. An odd number of slice moves since the cube was last in it's solved state causes OLL parity.
7. This Is a so called A9 commutator. It consists of: Setup-move, (A, B) Setup-move, where the first setup-move cancels one move with the first move of the A-part (R2 R becomes R'). A consists of R F R' and B consists of B2. 
What you do with the A-part is to exhange two corners without messing up the layer where B is done. You then exchange two corners with the B-move. Then you move back the piece that you first exchanged with A by doing A' and by that bring the exchanged corner to the other spot. Then you exchange the two last corners with B again. (I find it hard to explain this but basically all corners have now switched place, and nothing else is messed up)
8. U' D F' E2 F E2 F' E' B' y' U D' F E2 F' E2 F E
9. 
a) U' D F' E2 F E2 F' E' B' y' U D' F E2 F' E2 F E B
b) U' D F' E2 F E2 F' E' B2 y' U D' F E2 F' E2 F E B2


Fun challenge!


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## Renslay (Oct 17, 2013)

Here is my answers. I don't know all of them, of course I can search them, but what is the fun in that?



Spoiler



1) Any permutation is valid*, but the orientation of 7 corners defines the orientation of the last one. Therefore, for a randomly assembled cube the chance its validity is 1/3.

*There is an algorithm for swapping 2 pieces, and also another one for rotating one corner CW and one or CCW.

2) None of the elements is fixed (unlike the centers of a 3x3x3), so, let's say DBL is fixed (rotate the cube so DBL is in its place correctly). Every permutation is valid, which means 7 corners has 7! different permutations. 6 of the corners can have 3 different orientations independently, the last corner's orientation depends on the other ones, see the previous law. Therefore, we have 7!*3^6 = 3674160 legal positions.

3) Laws: a) the orientation of the 12th edge is determined by the other edges, b) the orientation of the 8th corner is determined by the other corners, c) the parity of the permutation of the edges is the same as the corners (i.e., both permutation is even or both is odd). 12 edges = 12! permutations, 8 corners = 8! permutations, edge orientations = 2^11 (or (2^12)/2 because of Law a) ), corner orientations = 3^7 (or (3^8)/3 because of Law b) ), which means 8!*12!*2^11*3^7 possible cube states devided by 2 because of Law c) -> 8!*12!*2^11*3^7/2 ~= 4.3*10^19 possible cube states.

4) I don't know.

5) I don't know.

6) The edge permutation is odd. Which occurs when the scramble has even (odd) number of quater turns of the edges, and you solve the edges with odd (even) number of quater turns: odd + even = odd number of moves, which means the permutation of the edges is still odd. OLL parity fix contains odd number quater turns on the edges, "fixing" its parity to even.

7) R' F R' B2 R F' R' B2 R2 = Lw' - U - R' D2 R - U' - R' D2 R - Lw = Lw' [U,R' D2 R] Lw = [Lw':[U,R' D2 R]]. It is a so called commutator, the heart of the commutator permute UBR, UFR, and BLD. Lw is just for setup / conjugate (ULB instead of BLD).

8)
a) Move every down side (yellow) piece to the top except the center, relatively solved b) do a single U turn, c) move everything back (inverse of a) )
a1) R2 L2 // moves 6 pieces from down to top
a2) U2 F2 M2 F2 M2 = U2 F2 R2 L2 B2 R2 L2 does an edge swap: every yellow piece is on the top side "solved"
b) U = simulate a D move, only the center differs
c) inverse of a)
Full solution: R2 L2 U2 F2 R2 L2 B2 R2 L2 U R2 L2 B2 R2 L2 F2 U2 R2 L2.
This results a single D move.

9)
a1) bring L center to U face: M E M'
a2) U move rotate the L center CW
a3) M E' M redo a)
a4) U' rotate U center CCW and solves the rest of the cube. The resulted algorithm is a commutator: M E M' U M E' M' U'

b) Change U to U2 in the previous example: M E M' U2 M E' M' U2

c) Do a) to rotate U center CCW, L center CW, then apply d) on U center: CCW + double turn = CW

d) U R L U2 L' R' does a 2-swap on the pieces, therefore doing the algorithm twice solves the cube. However, the number of U turns is 6, which results a double turn on the U center.


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## cannon4747 (Oct 17, 2013)

When I say _possible _positions, I refer to the number of positions you can reach through disassembly. When I say _legal_ positions, I refer to positions reachable through turning puzzle.

When I say "a 2x2 (or 3x3) cube", I am referring to the 2x2x2 and 3x3x3 3-dimensional puzzles. 2x2 and 3x3 cubes do not exist.

Question #1) Read Ryan Heise's page on cube laws of the 3x3. Now, write out the cube laws for the 2x2x2.


Spoiler: Answer 1



The only piece on a 3x3 that shows on a 2x2 is the corner piece. The rules about edges do not apply. Only 1/3 of possible corner orientations are reachable with legal moves. 7 of the 8 corners on the 2x2 can be oriented independently, with the orientation of the 8th corner depending on the orientation of the other 7.


Question #2) Calculate the number of legal positions to a 2x2x2. Give a brief explanation for each term in your calculation. Simply providing a numerical answer to this question does not sufficiently answer the question.


Spoiler: Answer 2



*Total legal 2x2 positions: 88,179,840*
Gained by multiplying the number of corner orientations by number of corner permutations.

Number of possible corner orientations: 2187
This number is derived from the law for the 2x2, 7 of the corners can each have one of 3 orientations with the 7th only being able to have 1. Or alternatively, only 1/3 of possible orientations are legal. All possible orientations would be expressed by 3^8. (1/3) * 3^8 = 3^7. 3^7 = 2187

Number of possible corner permutations: 40320
All corner permutations are possible on a 2x2. There are 8 corners and eight slots for corners. The first corner has 8 possible locations to be placed. The second has 7. So far, (8*7) is the answer for the first two corners. Following this logic, you get (8*7*6*5*4*3*2*1). Which can be expressed with factorial 8 (8!).
8! = 40320


Question #3) Calculate the number of legal positions to a 3x3x3. Give a brief explanation for each term in your calculation. Simply providing a numerical answer to this question does not sufficiently answer the question.


Spoiler: Answer 3



*Total number of 3x3 positions excluding center orientations: 43,252,003,274,489,856,000*
Taken from multiplying the possible orientations and permutations of the edges and corners.

Total corner orientations: 2187
derived from 2x2

Total corner permutations: 40320
Derived from 2x2

Total edge permutations: 239,500,800
Since you can only swap even sets of edges, and there are 12 edges, only half of the possible edge permutations are legal. This is because 5 of the 6 pairs can be permuted independently with the 6th depending on the others. (12!)/2 = (479,001,600)/2 = 239,500,800

Total edge orientations: 2048
The total number of possible edge orientations is reached by 2^12. However, since edges can only be flipped in pairs, 11 can be flipped independently with the 12th depending on the orientation of the others. The number of legal orientations is 2^11 = 2048.


Question #4) Write out the "Cube Laws" for the pyraminx. The "Cube Laws" are a list of any restrictions the puzzle must satisfy to be in a "legal" state.


Spoiler: Answer 4



*Pyraminx law 1: Only half the edge permutations are possible.*
Like 3x3, only even numbers of sets of edges can be swapped. This means that effectively, if the pyraminx has 6 edges, and 3 edge pairs, only 2 of those edge pairs are independent. The final one depends on the permutations of the other. This makes only half the permutations possible.

*Pyraminx law 2: Only half the edge orientations are possible.*
Edges can only be flipped in pairs, so therefore 5 of the 6 edges can be flipped independently with the orientation of the 6th depending on the other 5.


Question #5) Calculate the number of legal positions to a pyraminx. Include the rotations of the outer tips in your calculation. As usual, give a brief explanation for each term in your calculation. Simply providing a numerical answer to this question does not sufficiently answer the question.


Spoiler: Answer 5



*Total legal positions for pyraminx: 75,582,720*
Found by multiplying the possible positions and orientation for all the types of pieces.

Possible/legal orientations of tips: 81
There are four tips that rotate independently of each other and any other mechanism, they each have 3 possible orientations. That makes 3*3*3*3 = 3^4 = 81 possible orientations.

Possible/legal orientations of centers: 81
Experimenting with the algorithm R' U R U R' U R U and then rotating the tip U' shows that the centers can be rotated independently of anything else. The same exact math applies to the centers as the tips. 3*3*3*3 = 3^4 = 81

Legal orientations of edges: 32
There are 6 edges, and following pyraminx rule 2, only 5 of them can be flipped rotated independently with the orientation of the 6th depending on the other 5. Each edge has 2 possible orientations. This makes the equation: 2*2*2*2*2 = 2^5 = 32

Legal permutations of edges: 360
There are 6 edges and 6 slots for edges. According to rule 1, the permutation of the final 2 edges depends on the locations of the first 4. There are 6 slots for the first edge, 5 for the second, 4 for the third, and finally 3 for the fourth. This simplifies to the following equation: 6*5*4*3 = (6*5*4*3*2)/2 = (6!)/2 = (720)/2 = 360


Question #6) When solving the 4x4x4 using the standard reduction method (i.e. solve center groups, then pair up edge groups, then solve the puzzle as a 3x3x3), what is the cause of OLL parity? (Hint: It does not just "happen", you created it. Tell me how.)


Spoiler: Answer 6



The 4x4 "edge groups" or "wing edges" are actually symmetrical pieces. They can be swapped with one another using various algorithms. What happens on the 4x4 Oll parity is that the solver accidentally paired a set of these wing edges in the wrong permutation. It actually isn't an orientation parity at all, just another possible permutation for these types of pieces. If you look at the same phenomenon on a 5x5, one would see that the entire edge isn't flipped, but just those two outer edge pieces.


Question #7) Explain why the speedsolving A-perm alg R' F R' B2 R F' R' B2 R2 works. You can answer this question in two ways. Either A) Explain how this algorithm changes the positions of the affected corners, then as a separate part explain how it preserves their orientation, or B) explain the structure of this algorithm and classify it for what type of algorithm it is.


Spoiler: Answer 7



I use the names of the sides adjacent to pieces to say which piece I'm talking about. The UF edge is the edge in between the U-side and F-side centers. The same applies for corners.

This algorithm is based on the algorithm F R' B2 R F' R' B2 R which cycles 3 corners (UBL, UFR, DFR) in a clockwise fashion when viewed from the viewpoint of looking directly at the LF edge. This algorithm uses the setup move (or conjugate) R': so that the algorithm will rotate the UFR and the UBR corners with the UBL corner by moving them into the proper location.

The base algorithm is a commutator which allows the user to cycle all four F side corners with the UBL corner. From this point on, I am now rotating the algorithm as if I did an X cube rotation, the F face is now the U face, the D face is now the B face. The setup for the commutator is R' D2 R. Rotating the U face and then undoing the setup for the commutator will swap the UFR corner, the DBL corner, and whatever corner replaced the UFR corner when you did the U turn. After undoing the setup for the Commutator you may need to AUF to restore the orientation of the puzzle. You can use this algorithm to safely "store" U layer corners in the DBL slot and swap them with each other. Effectively, this can lead to another A perm algorithm: (R' D2 R) U' (R' D2 R) U' (R' D2 R) U2 (R' D2 R).


Question #8) Come up with a 5 generator algorithm on 3x3x3 that is different from my signature and explain it how it works. A 5-generator is defined as an algorithm that will use 5 sides of the cube to simulate a turn of the 6th side. Try the algorithm in my signature on a solved cube to get an idea of what this looks like.


Spoiler: Answer 8



*My goal move is a D2: R2 L2 (U2) R2 L2 F2 B2 (R2 U2)*3 F2 B2 (L2 U2)*3*

Inside of the parentheses are the actual algorithms I used. Setup moves are outside the parentheses. I used the algorithm (R2 U2)*3 in order to swap edges after I had swapped most of the layer with the R2 L2 U2.


Question #9) Come up with an algorithm to rotate two centers on a 3x3x3 supercube (if you don't have a supercube, make one!):
a) one 90 degrees clockwise, the other 90 degrees counter-clockwise
b) both 180 degrees
c) both 90 degrees clockwise


Spoiler:  Answer 9



*a: F E M' E' M F' M' E M E'*
Front center clockwise, Right center ccw.

*b: F2 E M' E' M F2 M' E M E' *
Front center and right center.

*c: F E M' E' M F' x (R U R' U)*5 x' M' E M E'*
Front and right centers clockwise.


Question #10) Come up with an algorithm to rotate exactly one center on a 3x3x3 supercube by 180 degrees.


Spoiler: Answer 10



*(R U R'U)*5*



That was fun. Took about an hour or two. I'd gladly enjoy more challenges like this. And also, seeing what other commutators and conjugates people can come up with is interesting.


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## Carrot (Oct 17, 2013)

Just gave it a quick go



Spoiler



1)
Every turn is an odd permutation, making every permutation legal.
Due to the corners being similar to 3x3x3 corners we also have only one third of the corner orientations reachable.
All in all, a third of the conceivable cube states are reachable by legal moves.

2)
There are 8 corners with 3 different orientations 3^8
There are 8 corners with 8 different permuations 8!
only 1/3 cases are reachable by legal moves
2x2x2 positions= 3^8*8!/3

3)
There are 8 corners with 3 different orientations 3^8
There are 8 corners with 8 different permuations 8!
There are 12 edges with 2 different orientations 2^12
There are 12 corners with 12 different permuations 12!
only 1/12 cases are reachable by legal moves
3x3x3 positions= (8^3*8!*2^12*12!)/12

4)
Every tip state can be obtained
Every center state can be obtained
Edges are an even permutations of edges, making half of them reachable
1/2 edge orientations can be obtained

5)
There are 4 tips with 3 different orientations 3^4
There are 4 centers with 3 different orientations 3^4
There are 6 edges with 2 different orientations 2^6
There are 6 edges with 6 different permutations 6!
only 1/4 cases are reachable by legal moves
pyraminx positions= ((3^4)^2*2^6*6!)/4

6)
a 4x4x4 OLL parity means that you have a 2-edge swap of an "edge", this is possible because when you do a wide turn you do 3 odd cycles, and therefore you can obtain odd cycles for edges and centers, however, you cannot see the cycles of the centerpieces as they all look the same for the same side, short story made long, the centers have an odd cycle. if centers are solved correct on a supercube you cannot get OLL parity.

7)
A-perm is a commutator of the form: [R2: [R F R', B2]] works by commutator magic

8)
You decided to steal the optimal one, so I will do this: L2 F2 B2 R2 F2 B2 U2 F2 B2 ->U<- B2 F2 U2 B2 F2 R2 B2 F2 L2

9)
YEW, I don't like supercube algs! xD


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## Renslay (Oct 17, 2013)

Hah, I just constructed an aswer for 8) which is different from yours but has the same length:



Spoiler



Mine:
R L' F2 B2 L' R U R' L B2 F2 L R' 

Yours:
R *L* F2 B2 L' *R'* U *R* L B2 F2 *L'* R'



It is different, right?


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## Jaycee (Oct 17, 2013)

Spoiler



1) All permutations are reachable. One third of orientations are reachable, because the orientation of a corner depends on the orientations of the other seven corners (and there are 3 possible orientations for a single corner). Therefore, 1/3 of all possible states of a randomly assembled 2x2x2 are reachable.

2) 3^7 * 8! / 24 = *3,674,160*. (3^7) represents the 3 possible orientations of 7 corners. The number is 3^7 and not 3^8 because the orientation of the final corner is dependent on the orientations of all other corners. The other term is (8!), which represents the legal permutations of the pieces. One corner can be in any of 8 possible places, another corner in any of the remaining 7 places, the next in any of the remaining 6, and so on. This number is divided by 24 because you can rotate the cube into 24 different positions and solve the corners the same way, due to lack of center pieces.

3)Corners: 3^7 * 8! = 88,179,840. (3^7) represents legal corner orientations. (8!) represents legal corner permutations.
Edges: 2^11 * 12! / 2 = 490,497,638,400. (2^11) is the number of legal collective edge orientations. This isn't 2^12 because the orientation of the 12th edge relies on the preceding 11 edge orientations. (12!) is the number of legal edge permutations. See explanation in the answer to question 2. The same idea is applicable to edges.
88179840 * 490497638400 = *43,252,003,274,498,856,000*. 

4) I'm a bit too tired to think about this one right now. I'll come back to it, probably.

5) 3^4 * 3^4 * 2^5 * 6! / 2 = *75,582,720*
Tips: Each of the four tips has three possible orientations (permutation is fixed). Therefore, 3^4 represents tips.
Centers: Centers technically are independent pieces (as shown by performing the alg R' U R U R' U R U and then doing u'. If you treat each piece that has three center stickers on it as a singular center piece, then there are 4 center pieces. Each piece can be rotated in 3 ways, and the permutation is fixed. Another 3^4 represents this for centers.
Edges: The number of legal edge orientations is 2^5. Each edge can be flipped in two ways, and the flip of one edge is decided by the orientation of the other 5 edges. The number of permutations can be described as 6!. One edge can be in any of the 6 spots, the next edge in any of the remaining 5 spots, and so on, until you reach the final 2 edges. 2 edges can not be swapped by themselves (another two would have to be swapped, or three edges would have to be cycled). In other words, the permutation of the final _*two*_edges are dependent on that of the other four.



I'm done for tonight.


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## cubernya (Oct 17, 2013)

Renslay said:


> Hah, I just constructed an aswer for 8) which is different from yours but has the same length:
> 
> 
> 
> ...



I would say no, since it's an inverse (I think, didn't look too closely).
I'll take a look at this tonight, so it'll be interesting. I think I can answer all but 8 (which is why I answered the above question)


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## Lchu613 (Oct 17, 2013)

theZcuber said:


> I would say no, since it's an inverse (I think, didn't look too closely).
> I'll take a look at this tonight, so it'll be interesting. I think I can answer all but 8 (which is why I answered the above question)


It's not an inverse.


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## Renslay (Oct 17, 2013)

theZcuber said:


> I would say no, since it's an inverse (I think, didn't look too closely).
> I'll take a look at this tonight, so it'll be interesting. I think I can answer all but 8 (which is why I answered the above question)



It's not an inverse, and basically both of them is in the form of A U A', but the two As are different (not just the movements themselves, but the resulted cubes too).

Jaycee, you feel right, your answer to 2) is incorrect.


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## Stefan (Oct 17, 2013)

Spoiler: Answer #8



Based on cannon4747's keen observation that you didn't request a quarter turn: *R2 F2 B2 L2 U2 L2 B2 F2 R2*. Setup, U2, setdown.
Quite nice written down as [R2 F2 B2 L2: U2] = D2, since every side appears exactly once.


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## Ranzha (Oct 18, 2013)

1.


Spoiler



All permutations are reachable. Only one-third of the orientations are reachable.



2.


Spoiler



7!*3^6/ = *3674160*.
Process:


Spoiler



Isomorphism:
The first order of business is to account for isomorphism. We don't want any of that business here. Thus, consider methodical assembly of a 2x2, where one corner is placed in each of the eight possible positions. Then consider that each corner's orientation can be defined in either of three ways. If we fix one corner (let's say DBL, as it is with scrambling conventions), we eliminate all possible isomorphism that could be attained by rotation of the puzzle.

Permutations:
Given a fixed corner, 7! is the number of possible/reachable permutations. If DBL is the first corner to be "assembled" as solved, this leaves any of the seven corners for the second position, six for the third, ad nauseam.

Orientations:
3^6 is the number of reachable orientations. Each corner can be oriented in any of three fashions except for the first corner, which has a fixed orientation, and the eighth corner, which has a defined orientation based on the complement to the modulus of the sum of the orientations of the other seven corners with respect to 3 (also hence the answer to number 1).






3.


Spoiler



*That 43 quintillion mumbo jumbo.*
Process:


Spoiler



Isomorphism:
Isomorphism isn't that big of a problem with the 3x3x3 as compared to the 2x2x2 because of the lack of visible core for the 2x2x2. The 3x3x3 isomorphic fix is attained with a fixed orientation of the now visible centres so that only face turns are legal moves.

On to the number crunching!

Same stuff as before, but this time we have to account for permutation parity as well as edge orientaiton parity. These will be sprinkled in with the respective calculations (CO, EO, CP, EP).

Corner Orientation:
As before, each corner can be oriented in any of three fashions. However, this time, instead of two corners being defined (remember, one of those two was used last time to accommodate for isomorphism), only one corner (the eighth corner) is defined. Thus, the calculation for corner orientation is *3^7*.

Edge Orientation:
Each edge can be oriented in either of two fashions excepting the twelfth edge, which has an orientation defined by the complement to the modulus of the sum of the orientations of the other seven corners with respect to 2. Thus, the calculation for edge orientation is *2^11*.

Corner Permutation:
Any of the eight corners can be placed in any of the corner positions. Permutation parity will be accounted for in the next calculation. Thus, the calculation for corner permutation (as before, save for isomorphism fix) is *8!*.

Edge Permutation:
Any of the twelve edges can be placed in any of the edge positions, save for the last two, whose permutation is defined a la the orientation commodities, except with regard to pairs of 2-swaps. Thus, the calculation for edge permutation accounting for permutation parity is *12!/2*.

Thus:
3^7 * 2^11 * 8! * 12! / 2 = *43252003274489856000*, aka "that 43 quintillion mumbo jumbo".






4.


Spoiler



All tip orientations are reachable. All three-faced centre orientations are reachable. Half of possible edge permutations are reachable. Half of possible edge orientations are reachable.



5.


Spoiler



3^4 * 3^4 * 2^5 * 6! / 2 = *75582720*.
Process:


Spoiler



Isomorphism:
A solved pyraminx can be oriented in twelve different ways. To accommodate for this, the centres/tips will have fixed permutations (which doesn't detract from the final calculation due to their physical permutations being fixed to the core in any case).

Tip orientation:
Each tip is independent and can be oriented in either of three fashions. Thus, the calculation for tip orientation is *3^4*.

Three-faced centre orientation:
Each three-faced centre is fixed to the core. Thus, the permutations of the three-faced centres are fixed. However, the orientations are independent of one another. Thus, the calculation for three-faced centre orientation is the same as it is for tip orientation: *3^4*.

Edge orientation:
As the state of two incorrectly oriented edges is reachable, but the state of one is not, the pyraminx is bounded by the restriction of edge orientation parity of order (or whatever it's called) 2. Thus, each of the edges can be oriented in any fashion except for the sixth edge, which has an orientation defined by the complement to the modulus of the sum of the orientations of the other seven corners with respect to 2. Thus, the calculation for edge orientation is *2^5*.

Edge permutation:
As the state of three incorrectly permuted edges is reachable, but the state of two is not, the pyraminx is bounded by the restriction of edge permutation parity of order (or whatever it's called) 2. Thus, edge of the edges can be permuted in any fashion except for the fifth and sixth edges, whose permutation is defined by the permutation of the other four edges. Thus, the calculation for edge permutation is *6!/2*.

Final calculation:
3^4 * 3^4 * 2^5 * 6! / 2 = *75582720*.






6.


Spoiler



"OLL parity" is caused by the edges being solved into an odd permutation as the centres are being solved. For this reason, all algorithms to solve OLL parity have an odd number of slice-altering moves (e.g. r, l, u).



7.


Spoiler



The A-perm is an A9 commutator. This means that it follows the outline of [A:[B, C]] while having one cancelling move between A and B or between C' and A'.
Premove (A): R2
Setup (B): R F R'
Exchange (C): B2

Thus, for this example:
[A:[B, C]] = A B C B' C' A'
= R2 R F R' B2 R F' R' B2 R2
= *R'* F R' B2 R F' R' B2 R2.

Here, the cancelling move is between A and B and NOT between C' and A'.



8.


Spoiler



L2 F2 B2 R2 D2 L2 B2 F2 R2.
Can be rewritten as L2 S2 L2 U2 R2 S2 R2. This leaves a U2.
Note that R2 S2 R2 = L2 S2 L2.
L2 S2 L2 offers a 2-2 swap of opposite edges and a 2-2 swap of opposite centres. Thus, the U2 move in the middle has no effect on the relative permutation of the edges nor the centres as the two U-layer edges that have been altered will maintain permutation relative to the other edges as well as to the U layer. Thus, the resulting R2 S2 R2 (acting the same way as L2 S2 L2) will leave only the U layer misaligned by the same amount as before, 180 degrees.



9.
a.


Spoiler



[M, E' L E]


b.


Spoiler



[M, E' L2 E]


c.


Spoiler



(R' U')63


d.


Spoiler



(U' R U' R')5


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## Jaycee (Oct 18, 2013)

Renslay said:


> Jaycee, you feel right, your answer to 2) is incorrect.



I think I figured it out.



Spoiler



I divided my original answer by 24, because there are 24 possible orientations of the cube itself (achieved by rotating, of course). I figured this out after about 8 minutes of looking and fiddling with a 2x2 and 3x3 in front of me and seeing how the cubes are different when solving the corners.

In short, I forgot to account for the fact that there are no center pieces.


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## elrog (Oct 18, 2013)

Cool quiz. I am assuming there will be more (possibly more advanced) quizzes to come because you said this is quiz # 1.


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## Renslay (Oct 18, 2013)

Stefan said:


> Spoiler: Answer #8
> 
> 
> 
> ...



But that is only a 5-generator on a subgroup, isn't it?


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## Stefan (Oct 18, 2013)

Renslay said:


> But that is only a 5-generator on a subgroup, isn't it?



Why "only"? Isn't it (also) one in the whole group?


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## Renslay (Oct 18, 2013)

Stefan said:


> Why "only"? Isn't it (also) one in the whole group?



I don't really understand your question*, but you have a point: your solution is indeed a solution according to the given defiition: "A 5-generator is defined as an algorithm that will use 5 sides of the cube to simulate a turn of the 6th side." Although, I think that's just a flaw in the definition rather than a loophole. Chris?

Edit: *Okay, maybe I chose poor words again. I wanted to ask how do you do a D turn using only the other 5 sides? With simulating D can simulate D2 and D', but with simulating D2 you can's simulate D or D'.


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## cmhardw (Oct 18, 2013)

elrog said:


> Cool quiz. I am assuming there will be more (possibly more advanced) quizzes to come because you said this is quiz # 1.



I have enough questions for probably 2 more quizzes. If anyone has a cool theory idea related to twisty puzzles that they would like to see on a future quiz, feel free to PM me. I don't really intend these to be quizzes in the usual sense. I intend these to be threads where people learn cool things about twisty puzzles that they maybe previously were unaware of 



Renslay said:


> I don't really understand your question*, but you have a point: your solution is indeed a solution according to the given defiition: "A 5-generator is defined as an algorithm that will use 5 sides of the cube to simulate a turn of the 6th side." Although, I think that's just a flaw in the definition rather than a loophole. Chris?



Stefan pointed out that I have a flawed definition of a 5-generator algorithm. I will happily accept any alg that performs a quarter turn or a double turn on the 6th side, as I don't want to edit questions too much after people have already started answering.

Also Renslay, your answer to number 8 looks fine to me


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## Stefan (Oct 18, 2013)

cmhardw said:


> 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.


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## Rune (Oct 18, 2013)

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)


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## cmhardw (Oct 18, 2013)

Stefan said:


> 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?


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## porkynator (Oct 18, 2013)

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).


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## Stefan (Oct 18, 2013)

Rune said:


> 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.



cmhardw said:


> 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"


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## Rune (Oct 18, 2013)

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


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## Stefan (Oct 18, 2013)

Rune said:


> 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."_


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## Carrot (Oct 18, 2013)

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'


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## cuBerBruce (Oct 19, 2013)

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>.)


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## elrog (Oct 19, 2013)

^
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.


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## porkynator (Oct 19, 2013)

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


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## Rune (Oct 19, 2013)

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".


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## Renslay (Oct 19, 2013)

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


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## Stefan (Oct 19, 2013)

cuBerBruce said:


> 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.


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## Ranzha (Dec 5, 2013)

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


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