Random Cubing Discussion

[Cale S]

What is the shortest (or a short) sequence of moves Q, such that every position can be solved by crossing out some moves of Q and then executing the rest in the original order?

For instance: say your sequence is R U R2, then you could execute nothing or R or U or R2 or R U or R R2 (= R') or U R2 or R U R2. But you can't do U U or R2 U.

There is an obvious theoretical lower limit of 66 moves (with fewer, there are not enough subsets of Q's moves) but I think the real minimum will be quite a bit more than that. We could also create an upper bound by defining a method that solves one piece at a time and just writing down all the solutions in order, i.e. start with a list of algs to solve DB from every position, write them in order, then get a list of algs to solve DF from every position without disturbing DB, write those in order at the end, etc. etc. I don't know how big this would actually be.

If you allow rotations but count them as moves, the shortest I can find is
(y x x2 U U2)20
100 etm

Using wide moves instead of rotations:
(Uw D' Rw L' Rw2 Lw2 U U2)20
160 htm

If you only use normal face turns, I can't think of anything better than what Stefan said.

 

[obelisk477]

Thinking about getting these fingerweights for Christmas to improve TPS

 

[mark49152]

Not sure about posts per day...

Looks like a shoot-out between you and Kennan

 

[guysensei1]

Looks like a shoot-out between you and Kennan

I'm lagging behind by 1 post per day.

(I'd say that I have the unofficial AsR for posts per day...)


EDIT: also, Kennan wins. He's been here for less time (few days less) and he has ~200-300 more posts

 

[TDM]

I'm lagging behind by 1 post per day.

(I'd say that I have the unofficial AsR for posts per day...)


EDIT: also, Kennan wins. He's been here for less time (few days less) and he has ~200-300 more posts

I didn't post for around two months though - I first posted in may. I also didn't really post as much when I first started - is there any way to calculate posts/day for just 2014?

 

[mark49152]

I didn't post for around two months though - I first posted in may. I also didn't really post as much when I first started - is there any way to calculate posts/day for just 2014?

If you really want to maximise your score, why not just spam the forum more?

 

[guysensei1]

D L B2 L F2 R2 B2 R' U2 B D' L' F L' D2 B L' D2 F2
D' L' B2 L' F2 R2 B2 R U2 B' D L F' L D2 B' L D2 F2

Take a look at these 2 scrambles. I've simply inverted every individual move of the scramble. Apart from the visual similarity, are these scrambles related in any way? Can one predict what the second scramble will be like given a picture of what the first scramble looks like?

Copied this over from OAQT. It wasn't gonna get answered there and will probably get buried.

 

[goodatthis]

Copied this over from OAQT. It wasn't gonna get answered there and will probably get buried.

I really don't know about scrambkes like that, but I do know that inverse scrambles work something like this:

Normal: Block A needs to go in Position A, currently in Position B
Inverse: Block B needs to go in Position B, currently in Position A
So basically whatever position a piece occupies, the piece that needs to go there will occupy the solved position of the original piece on the inverse scramble.

 

[whauk]

Copied this over from OAQT. It wasn't gonna get answered there and will probably get buried.

If you execute 2 T-perms you get a solved cube. However if you execute R' U' R U R F' R2 U R U R' U' R F R' U' R U R F' R2 U R U R' U' R F (that is 2 T-perms with every move inverted) you get some pretty arbitrary looking position. So you cannot identify the other position. Also the two move sequences don't fix the same pieces. I assume it is pretty safe to say, that "inverting every move" has no useful properties.

 

[Cale S]

D L B2 L F2 R2 B2 R' U2 B D' L' F L' D2 B L' D2 F2
D' L' B2 L' F2 R2 B2 R U2 B' D L F' L D2 B' L D2 F2

Take a look at these 2 scrambles. I've simply inverted every individual move of the scramble. Apart from the visual similarity, are these scrambles related in any way? Can one predict what the second scramble will be like given a picture of what the first scramble looks like?

If you were only given an image of the first scramble, there are infinitely many ways you could get to that position (and probably multiple ways that would be move-optimal). If you see that doing D L B2 L F2 R2 B2 R' U2 B D' L' F L' D2 B L' D2 F2 will get to that position and invert every move, you will get this. However, you could also reach the first position with L' B' D2 L F' R' D R L U' R2 F2 U2 R2 D F2 D2 R2 D' F2 B2 L2, and inverting every move of that sequence of moves gives you something different. Unlike inverting a scramble, inverting each individual move doesn't correlate two specific scrambles.

Spoiler: explanation of why it's so random

Every face turn you can do either does four 2-cycles (2 of corners, 2 of edges) or two 4-cycles (1 of corners, 1 of edges). Let's say we have the scramble D F. If we focus on only the edges, the D move does the 4-cycle DF>DR>DB>DL. Then the F move does the 4-cycle DF>LF>UF>RF. The final result is the same as doing the 7-cycle DF>DR>DB>DL>LF>UF>RF.
If you invert each move, the 4-cycles become inverted, so D' does DF>DL>DB>DR and F' does DF>RF>UF>LF. The result is the 7-cycle DF>DL>DB>DR>RF>UF>LF. If we say that the 7-cycle the original scramble did was a>b>c>d>e>f>g, the scramble with each move inverted would be a>d>c>b>g>f>e. The order the pieces are cycled in was completely messed up, and that was only with 2 moves. With up to 20 moves, the cycles are rearranged so much that there is no correlation whatsoever. It would be like doing BLD but doing your memo in a random order. The result is entirely random.

 

[Stefan]

Spoiler: explanation of why it's so random

Every face turn you can do either does four 2-cycles (2 of corners, 2 of edges) or two 4-cycles (1 of corners, 1 of edges). Let's say we have the scramble D F. If we focus on only the edges, the D move does the 4-cycle DF>DR>DB>DL. Then the F move does the 4-cycle DF>LF>UF>RF. The final result is the same as doing the 7-cycle DF>DR>DB>DL>LF>UF>RF.
If you invert each move, the 4-cycles become inverted, so D' does DF>DL>DB>DR and F' does DF>RF>UF>LF. The result is the 7-cycle DF>DL>DB>DR>RF>UF>LF. If we say that the 7-cycle the original scramble did was a>b>c>d>e>f>g, the scramble with each move inverted would be a>d>c>b>g>f>e. The order the pieces are cycled in was completely messed up, and that was only with 2 moves. With up to 20 moves, the cycles are rearranged so much that there is no correlation whatsoever. It would be like doing BLD but doing your memo in a random order. The result is entirely random.

That doesn't explain it.

Spoiler: Hint

You can't possibly explain it with only two sides, because in those cases, there is a strong connection and it isn't random.

 

[TDM]

EDIT: also, Kennan wins.

OR DOES HE!?

 

[guysensei1]

OR DOES HE!?

OMGWTFBBQ IS THIS sup-10 posts per day?!


EDIT: also, how often is the posts per day count updated?

 

[TDM]

OMGWTFBBQ IS THIS sup-10 posts per day?!


EDIT: also, how often is the posts per day count updated?

It was 11.8x at the time of posting He'd joined less than two days ago though.
and I don't know... I think he had 11.87 posts/day, which suggests that it could be updated constantly, because no integer divided by 2 gives you a decimal ending in .87.

 

[mark49152]

It was 11.8x at the time of posting He'd joined less than two days ago though.
and I don't know... I think he had 11.87 posts/day, which suggests that it could be updated constantly, because no integer divided by 2 gives you a decimal ending in .87.

Give him a chance, two days is not a fair sample. I bet there is a two-day window somewhere in your membership where you way exceeded 11.87

 

[guysensei1]

Give him a chance, two days is not a fair sample. I bet there is a two-day window somewhere in your membership where you way exceeded 11.87

TDM messaged me about posting 40 posts in 3 hours or something like that some time ago XD

EDIT: It was 17 posts in 2 hours. So 17/2*24= 204 posts per day XD

 

[TDM]

Give him a chance, two days is not a fair sample. I bet there is a two-day window somewhere in your membership where you way exceeded 11.87

I know 2 days isn't a fair sample, that's what I was trying to say... and I'm sure there has been more than one day where I've gone over 20 posts

 

[mark49152]

TDM messaged me about posting 40 posts in 3 hours or something like that some time ago XD

EDIT: It was 17 posts in 2 hours. So 17/2*24= 204 posts per day XD

For a split second there, you were at an almost infinite number of posts per day

 

[guysensei1]

When can a 1LLL case be solvable by F <R,U,D> F'?

 

[IRNjuggle28]

When can a 1LLL case be solvable by F <R,U,D> F'?

My guess is any permutation with at least 2 edges oriented. <R,U,D> covers any PLL case and any OCLL case, and F (RUD) F' can orient either 2 or 0 edges, but not 4. It might depend on whether you allow cube rotations or AUF before executing the alg. Being able to choose which face is F would matter for orienting edges, if not for other stuff too. And maybe having all the RUD moves fall between the F and the F' limits what can be done in a way I haven't thought of. Someone who knows puzzle theory confirm pls.