Floppy Cube questions

Robert-Y · Apr 15, 2009

1. What is greatest number of moves needed to solve a floppy cube?

2. In order to solve one in one algorithm, how many algs would you need to learn?

byu · Apr 15, 2009

96 for the second one
For the first, it's 12 for my method, but I sure it's lower optimally

Johannes91 · Apr 15, 2009

Robert-Y said:
1. What is greatest number of moves needed to solve a floppy cube?

Eight.

Code:

Depth: 0	Positions: 1	1
Depth: 1	Positions: 5	4
Depth: 2	Positions: 15	10
Depth: 3	Positions: 39	24
Depth: 4	Positions: 92	53
Depth: 5	Positions: 156	64
Depth: 6	Positions: 187	31
Depth: 7	Positions: 191	4
Depth: 8	Positions: 192	1

Average: 4.427083333333333

The 192 can be calculated as (4! * 2^4) / 2.

The only case that requires 8 can be solved with e.g. F2 B2 R2 F2 R2 L2 F2 R2.

FYI: I defined floppy cube as a 3x3x3 where UF,UB,UR,UL,FR,FL need to be solved and the only allowed moves are F2,B2,R2,L2. I first considered using a super 3x3x3 where E slice needs to be solved, which would've been much cleaner, but I haven't tested the solver with super cubes yet.

Robert-Y said:
2. In order to solve one in one algorithm, how many algs would you need to learn?

Depends on what counts as an alg. Just a few non-trivial ones should suffice.

Robert-Y · Apr 15, 2009

Thanks Johannes

Well what I really meant was: how many sequences would you have to learn in order to solve the floppy cube in the maximum number of moves required every time just by looking at it once? That is, so that you could even do the floppy cube blindfolded in the maximum number of moves required.

Johannes91 · Apr 15, 2009

Yes, I think I understood you, but would you count 3-move sequences as algorithms that have to be learned?

I counted manually, and there seem to be about 35 unique cases. Only one requires 8 and one requires 7. I don't have a real floppy cube but after doing some <F2,B2,R2,L2> solves on a 3x3x3, it feels like a really easy puzzle to master. Seeing 5-6 move solutions with so few pieces isn't hard.

If you really want to learn an alg for every case, then you need 35 or 192 or something in between, depending on how much you take advantage of mirrors/AUF/inverses.

*LukeMayn* · Apr 15, 2009

I guess the recognition could be considered like "recognizing" how to do the cross.

byu · Apr 15, 2009

No, it's totally different. Because a cross is just 4 edge pieces that have no effect on each other. On a floppy cube we have 8 pieces that are dependent (you can't swap two corners without flipping an edge)

And Robert, I average sub-7 BLD with floppy cube just by speed blinding it. It's not that hard

JohnnyA · Apr 15, 2009

What IS a floppy cube?

jason9000 · Apr 15, 2009

A 1x3x3.
http://www.google.com/search?q="floppy+cube"

Johannes91 · Apr 15, 2009

I think it's much easier than cross (or 2x2x2 block or EO line). But similar in the sense that all the optimal solutions are understandable.

JohnnyA said:
What IS a floppy cube?

http://www.geocities.com/beksendeavors/images/floppy_cube.gif

Stefan · Apr 15, 2009

Wud's teh algirythm for just flippin the center upside down? (double dot pattern)

cpt.Justice · Apr 15, 2009

StefanPochmann said:
Wud's teh algirythm for just flippin the center upside down? (double dot pattern)

Here's a clue: It can be written as a+bi where b≠0

Or can it?!

The field of imaginary algorithms is one that needs to be explored.

Edit:

Dictionary.com defines an algorithm as "a set of rules for solving a problem in a finite number of steps".

Code:

[B]step 1:[/B]
a=0
while a==0:
   if you have a floppy cube:
      a=1
   else:
      get a floppy cube.
end
proceed to step 2
[B]step 2:[/B]
a=0
while a==0:
   if you have hands:
      remove the stickers on the centers.
      a=1
   else:
      b=0
      while b==0:
         if you have friends with hands:
            have one of your friends with hands remove the stickers on the centers.
            a=1
         else:
            get a friend with hands.
            b=1
      end
end
proceed to step 3.
[B]step 3:[/B]
put the previously removed stickers on the opposite side of the floppy cube yourself OR
have your friend with hands put the previously removed stickers on the opposite side of the floppy cube.

Stefan · Apr 15, 2009

OMG you have way too much time. Was funny, though. Then again, I'm still waiting for proof of the impossibility.

I'd also like to know how often this strategy produces suboptimal solutions and how suboptimal they are:
1. Optimally solve the 2x2 block that has the shortest solution.
2. Optimally solve the rest turning only the two sides opposite that first block.

Tim Reynolds · Apr 15, 2009

Wait, if you don't have hands, you get stuck in an infinite loop in the "while b==0" part--once you get a friend with hands, you keep repeating that loop over and over!

Step 3 is confusing...I don't know which option to choose...

This may be incorrect as I don't own a floppy cube, but I don't believe that that position is possible...you can do a pons asinorum, but you would then have to basically swap the top and bottom stickers on the corners, which is just physically impossible.

Johannes91 · Apr 15, 2009

StefanPochmann said:
1. Optimally solve the 2x2 block that has the shortest solution.
2. Optimally solve the rest turning only the two sides opposite that first block.

What if there are multiple 2x2 blocks that have the shortest solution? Pick one randomly or pick the one that gives best step 2?

Tim Reynolds said:
This may be incorrect as I don't own a floppy cube, but I don't believe that that position is possible...

Sounds correct to me.

Stefan · Apr 15, 2009

Johannes91 said:
StefanPochmann said:

1. Optimally solve the 2x2 block that has the shortest solution.
2. Optimally solve the rest turning only the two sides opposite that first block.

Click to expand...

What if there are multiple 2x2 blocks that have the shortest solution? Pick one randomly or pick the one that gives best step 2?

Sorry about that. As I'm interested how much this can fail, always pick the one giving the *worst* step 2.

Btw, I did find a four moves case that this strategy solves in six moves: L U R U. Without thinking it through, I suspect this is the shortest failure case.

Robert-Y · Apr 16, 2009

For the proof of impossibility of "flipping the centre upside down": That involves permuting edges which we know is impossible, is that good enough?

byu · Apr 16, 2009

L2 R2 B2 L2 R2 F2 L2 R2

That's the dot

Johannes91 · Apr 16, 2009

The "pessimistic" method below always chooses the 2x2 block and optimal solution for it that gives the worst Step 2, and the "optimistic" method does the opposite.

Code:

          | Pessimistic | Optimistic |  Optimal
----------+-------------+------------+-----------
Depth: 0  |    1   1    |    1   1   |    1   1
Depth: 1  |    5   4    |    5   4   |    5   4
Depth: 2  |   15  10    |   15  10   |   15  10
Depth: 3  |   39  24    |   39  24   |   39  24
Depth: 4  |   68  29    |   76  37   |   92  53
Depth: 5  |  100  32    |  120  44   |  156  64
Depth: 6  |  145  45    |  165  47   |  187  31
Depth: 7  |  181  36    |  189  24   |  191   4
Depth: 8  |  192  11    |  192   1   |  192   1
----------+-------------+------------+-----------
Average:  | 5.11458333  | 4.80208333 | 4.42708333

Cases where pessimistic uses more moves than optimal:

Code:

Pessimistic              Optimistic             Optimal

8f: B L - F R F R F R    4f: R B - L F          4f: R B L F
6f: B - F L F L F        6f: B - F L F L F      4f: R F R B
6f: B L - R F R F        6f: B L - R F R F      4f: R B R L
7f: F L - B R B R B      5f: L B - L F L        5f: R F L B R
7f: F L F - B L B L      5f: R B R - L B        5f: R F R L F
7f: F R - B L B L B      5f: R B - R F R        5f: R B R F R
7f: B - F R F R F R      7f: B - F R F R F R    5f: R B L F L
7f: B - L F L F L F      7f: B - L F L F L F    5f: R F R B L
7f: R F - B R B R B      7f: R F - B R B R B    5f: F L B R L
7f: R B R - L F L F      7f: R B R - L F L F    5f: R L B R L
8f: L F L F - B L B L    6f: R L B R - L B      6f: F R L F R L
8f: R F R F - B L B L    6f: F B L F - B L      6f: F B R F B R

byu said:
L2 R2 B2 L2 R2 F2 L2 R2

That's the dot

F2 R2 L2 B2

But that doesn't "just flip the center upside down".

Stefan · Apr 16, 2009

Thanks Johannes, very nice statistics. The bad cases are also interesting, especially the first, taking eight moves instead of four. I was hoping it'd never waste more than two moves. But at least it doesn't worsen the worst case to more than eight moves. What program did you use for this? Or did you write one just for this?

Do you have a real floppy cube? I hold mine rather vertically so that UDLR are my moves. Because I put my thumbs/fingers on the large faces and for the 3x3x3 my thumbs/fingers are on F/B. I can of course use FBLR, too, I'm just curious whether it comes from theory or practice.

Johannes91 said:
byu said:

L2 R2 B2 L2 R2 F2 L2 R2

That's the dot

Click to expand...

F2 R2 L2 B2

But that doesn't "just flip the center upside down".

Yes, the word "just" was very intentional.

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