Random Cubing Discussion

[Tim Major]

I wonder how much the 2x2 WR average would go up without inspection.

Everyone would just use CLL with inefficient layers, so TPS and CLL recognition would be the only things that matter. Would remove the solution from mattering entirely

 

[qqwref]

<R,U,y> cube-in-cube-in-cube alg:

R' U R' U' R2 U2 R U' R2 y2 U2 R2 U2 R' U' y R2 y' U R U' y R' U2 R2 U' R' y' R2 y2

 

[GG]

<R,U,y> cube-in-cube-in-cube alg:

R' U R' U' R2 U2 R U' R2 y2 U2 R2 U2 R' U' y R2 y' U R U' y R' U2 R2 U' R' y' R2 y2

bold y2 redundancy x) other than that nice alg. I prefer it like this :3
R' U R' U' R2 U2 R U' R2 U2 L2 U2 L' U y' R2 y' U R U' y R' U2 R2 U' R' y' R2

 

[IRNjuggle28]

How do I access my post history in specific threads? I'm looking for a list of every Accomplishment Thread post I've ever made, if someone could show me how to get that.

And my earlier question: How do I get a third signature line?

 

[Stefan]

How do I access my post history in specific threads? I'm looking for a list of every Accomplishment Thread post I've ever made, if someone could show me how to get that.

Go to the subforum where the thread is, click on the Replies number, then in the popup click on your Posts number.

Edit: Nice, Thom currently leads this thread with 2[sup]8[/sup] posts (I like powers of 2).

 

[Brest]

How do I get a third signature line?

You press enter, you are on a new line, and then type.

 

[guysensei1]

You press enter, you are on a new line, and then type.

You forgot the part about becoming a premium member.

 

[IRNjuggle28]

You press enter, you are on a new line, and then type.

lol

You forgot the part about becoming a premium member.

I'd wondered if that was it, but this didn't mention it.

 

[cmhardw]

I am attending a talk at Christopher Newport University today by one of the math professors who is giving an overview of the group theory topics that apply to the cube. His talk will cover the order of a permutation, and I am supplying a number of cubes for his participants to be able to attempt to experience the order of a permutation by applying an algorithm repeatedly to a solved cube until it solves again. I am present at the talk, in part, to help unscramble any wayward cubes that get messed up in that process I am also interested in the group theory topics, and seeing them presented in a proper way by a mathematician. Very cool!

 

[IRNjuggle28]

I am attending a talk at Christopher Newport University today by one of the math professors who is giving an overview of the group theory topics that apply to the cube. His talk will cover the order of a permutation, and I am supplying a number of cubes for his participants to be able to attempt to experience the order of a permutation by applying an algorithm repeatedly to a solved cube until it solves again. I am present at the talk, in part, to help unscramble any wayward cubes that get messed up in that process I am also interested in the group theory topics, and seeing them presented in a proper way by a mathematician. Very cool!

If a video is taken, you should post the link on here. Sounds interesting.

 

[Pailly19]

well, there are only 2 stores here that sell rubiks brands. one also sell V6 and V7
i asked them why the don't sell V5 and they told me it was because they already got rubiks 5X5:fp

 

[Ágoston Török]

Hello guys,
I have a question. I was wondering how many possible cases does the Last Layer have? So if I wanted to solve the LL with one algorithm always (just in theory ofc.) how many algorithms did I have to learn? (something tells me that the answer is not just simply 21*57 (+1 solved case).

 

[Randomno]

Hello guys,
I have a question. I was wondering how many possible cases does the Last Layer have? So if I wanted to solve the LL with one algorithm always (just in theory ofc.) how many algorithms did I have to learn? (something tells me that the answer is not just simply 21*57 (+1 solved case).

1221.

 

[TDM]

Hello guys,
I have a question. I was wondering how many possible cases does the Last Layer have? So if I wanted to solve the LL with one algorithm always (just in theory ofc.) how many algorithms did I have to learn? (something tells me that the answer is not just simply 21*57 (+1 solved case).

Including solved, 1212. There's a page on the wiki for this: 1lLL.

 

[goodatthis]

Hello guys,
I have a question. I was wondering how many possible cases does the Last Layer have? So if I wanted to solve the LL with one algorithm always (just in theory ofc.) how many algorithms did I have to learn? (something tells me that the answer is not just simply 21*57 (+1 solved case).

I'm pretty sure that this is how it's done

51 OLL cases with no symmetry, 51*72= 3672
5 OLL cases with U2 symmetry, 5*36= 180
2 OLL cases with 4-way symmetry, 2*21= 42

3672+180+42=3894 possible algs to learn. The 1212 figure excludes mirrors and inverses (I think, maybe something else too), so it's not really accurate. You might as well say that full ZBLL is 177 algs.

Easy way to figure out things like this:

So I got the 72, 36, and 21 by finding the number of possible PLL cases for each. Technically, PLL is 21 algs because the top face is solved, so you can simply perform an alg from a different angle because a solved face has what's known as symmetry.

A good way to think of symmetry is if you can do a U/U'/U2 and perform the alg (and it solves the case), the case has total symmetry, if you can only do a U2 the case has U2 symmetry, and if you can't do any, the case has no symmetry. Keep in mind that when I say case, I mean orientation cases and permutation cases separately.

So for PLL, there are 4 cases with total symmetry, (H, N, N, solved) 2 cases with U2 symmetry (Z and E) and 16 with no symmetry. From that, there's 4 cases for each no symmetry case, 2 cases for each U2 symmetry case, and 1 case for each total symmetry case. So (4*1)+(2*2)+(16*4)=72, so 72 total PLL cases.

Now if we do the same for OLL, we get 51 no symmetry cases, 5 U2 symmetry cases, and 2 total symmetry cases. For the U2 symmetry cases, we have to halve the number of possible PLL cases, so we get 36, and for the total symmetry cases (pure 4-flip, and solved), we simply take the number of PLL cases as we see them, so 21. Then we just do the math above, and theres the answer.

 

[Ágoston Török]

I'm pretty sure that this is how it's done

51 OLL cases with no symmetry, 51*72= 3672
5 OLL cases with U2 symmetry, 5*36= 180
2 OLL cases with 4-way symmetry, 2*21= 42

3672+180+42=3894 possible algs to learn. The 1212 figure excludes mirrors and inverses (I think, maybe something else too), so it's not really accurate. You might as well say that full ZBLL is 177 algs.

Easy way to figure out things like this:

So I got the 72, 36, and 21 by finding the number of possible PLL cases for each. Technically, PLL is 21 algs because the top face is solved, so you can simply perform an alg from a different angle because a solved face has what's known as symmetry.

A good way to think of symmetry is if you can do a U/U'/U2 and perform the alg (and it solves the case), the case has total symmetry, if you can only do a U2 the case has U2 symmetry, and if you can't do any, the case has no symmetry. Keep in mind that when I say case, I mean orientation cases and permutation cases separately.

So for PLL, there are 4 cases with total symmetry, (H, N, N, solved) 2 cases with U2 symmetry (Z and E) and 16 with no symmetry. From that, there's 4 cases for each no symmetry case, 2 cases for each U2 symmetry case, and 1 case for each total symmetry case. So (4*1)+(2*2)+(16*4)=72, so 72 total PLL cases.

Now if we do the same for OLL, we get 51 no symmetry cases, 5 U2 symmetry cases, and 2 total symmetry cases. For the U2 symmetry cases, we have to halve the number of possible PLL cases, so we get 36, and for the total symmetry cases (pure 4-flip, and solved), we simply take the number of PLL cases as we see them, so 21. Then we just do the math above, and theres the answer.

Thank you very much!

 

[guysensei1]

R U R' U F' U' F and R B' R B R' U R' are 7 move optimal solutions for the exact same position on the cube.As far as I can tell the 2 solution aren't related in any way. Are there any such pairs of optimal solutions that are 6 moves or less?

EDIT: I realise taking away the first R will give a 6 move pair.
U R' U F' U' F and B' R B R' U R'

 

[Stefan]

Those sixers are quite related. Invert B' R B R' U R' and you get R U' R B' R' B, then rotate the cube so that R and U switch places and you get the first alg.

Jaap has a similar triple here:
http://www.jaapsch.net/puzzles/cayley.htm#identities

 

[guysensei1]

Those sixers are quite related. Invert B' R B R' U R' and you get R U' R B' R' B, then rotate the cube so that R and U switch places and you get the first alg.

Thanks. I see how they're related now.

Jaap has a similar triple here:
http://www.jaapsch.net/puzzles/cayley.htm#identities

Hmm. Interesting page. God's number for 2 gen scrambling is still 20?! Example of such a position?

 

[TDM]

Hmm. Interesting page. God's number for 2 gen scrambling is still 20?!

Interesting... is God's number 20 for 3-gen (both RUL and RUF)?

(also, something else hopefully random enough for this thread, if you don't count the 67 days between my join date and my first post I have over 7 posts per day now: 3971/((3971/6.29)-67)=7.04, not including this post)