# 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?


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## 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


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## Johannes91 (Apr 15, 2009)

Robert-Y said:


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


Eight.


```
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.


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## 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.


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## 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.


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## *LukeMayn* (Apr 15, 2009)

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


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## 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


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## JohnnyA (Apr 15, 2009)

What IS a floppy cube?


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## jason9000 (Apr 15, 2009)

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


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## 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


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## Stefan (Apr 15, 2009)

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


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## 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".

```
[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.
```


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## 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.


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## 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.


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## 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.


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## Stefan (Apr 15, 2009)

Johannes91 said:


> StefanPochmann said:
> 
> 
> > 1. Optimally solve the 2x2 block that has the shortest solution.
> ...


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.


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## 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?


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## byu (Apr 16, 2009)

L2 R2 B2 L2 R2 F2 L2 R2

That's the dot


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## 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.


```
| 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:


```
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"_.


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## 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
> ...


Yes, the word "just" was very intentional.


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## Stefan (Apr 16, 2009)

Robert-Y said:


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



Do an M-move and you have permuted the edges.


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## Johannes91 (Apr 16, 2009)

StefanPochmann said:


> What program did you use for this? Or did you write one just for this?


I wrote the statistics part just for this, but for generating the solutions I used a program/library that makes solving partial 3x3x3s fairly easy. Deciding what I want to do exactly (formatting etc.) took longer than writing the code to do it.



StefanPochmann said:


> 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.


Nope, I don't have one. I chose FBRL because I usually visualize domino in that orientation, and I also use <U,D,F2,B2,R2,L2> in Kociemba's algorithm. Didn't really think about it; when executing the solutions on a 3x3x3, I actually do (half) an x' at the beginning.


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## cpt.Justice (Apr 16, 2009)

Tim Reynolds said:


> 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!



No you don't. The first while loop will end. Then you proceed to the next step.



Tim Reynolds said:


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



You can do either, or both. It's basic boolean logic.



StefanPochmann said:


> Was funny, though.



Yay. My aim was mildly amusing at best.


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## Robert-Y (Apr 16, 2009)

StefanPochmann said:


> Robert-Y said:
> 
> 
> > For the proof of impossibility of "flipping the centre upside down": That involves *permuting edges which we know is impossible*, is that good enough?
> ...



Ok what about... you can't permute edges without flipping the centre therefore flipping the centre alone is impossible?

This is a random question but is it possible to flip the centre on a hexagonal floppy prism?


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## Stefan (Apr 16, 2009)

Robert-Y said:


> you can't permute edges without flipping the centre


But I *want* to flip the center.



Robert-Y said:


> *therefore* flipping the centre alone is impossible


I fail to see that connection.


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## Robert-Y (Apr 16, 2009)

Grrr.... Erm...ok don't have anything right now *starts thinking harder*


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## TomZ (Apr 16, 2009)

Here is my over elaborate proof that you cannot turn what I like to refer to as the core rather than the center around on it's own:

A turn of any of the four faces of the floppy cube does the following:

- It changes the "orientation" of a center piece (also referred to by some as edge).

- It swaps two "corners", also turning both around.

The action of turning the corner around is irrelevant because the orientation of the corner is dependent on it's position. This is because if we consider one corner we can see that there are four different places it could go.

Two of these places can only be reached by doing an even number of turns, the other two places can only be reached by doing and odd number of turns.

Consider the FR corner. We can turn the right face which moves it to BR. We have now done an odd number of turns. If we continue doing turns that involve this corner and do not reverse our steps, we will find that the next time this corner is in BR we will have done four more turns. As four is an even number, and any even number multiplied by any other natural number is also an even number we find that to get this corner to this position we will always do an odd number (1+any even number) of turns involving this corner.

Because each turn flips the corners it permutes corner around, we can see that the positions reached by an odd number of turns will always have it's corner turned from the identity, and the positions reached by an even number of turns will not have the corner turned from the identity.

We do not need to consider middle layer turns as they are identical to turning the entire prism over and doing two non intersecting moves such as L R.

To flip the core piece (referred to some as center piece), we need to change the orientation of all the edge pieces. That would require an even number of face turns greater than or equal to 4.

The corners need to be at their identity positions. This requires an even number of face turns, and given that an even number of face turns is required to flip the center pieces, this is possible.

Next, the corners need to be flipped around. We have already proven that this is impossible without permuting them. It is therefor impossible to reach the 'dot' state. Q.E.D.


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## Stefan (Apr 16, 2009)

I'd say TomZ's proof is somewhat an overly elaborate version of Tim's sketch:


Tim Reynolds said:


> 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.


I'm not quite satisfied with either, but admittedly the reason I asked was that I couldn't come up with a nice concise way to explain/prove it myself.

Now I came up with this: Look at the left cube in the image below. Whatever moves you make, the three highlighted stickers will always stay fixed relative to each other. So it's not possible to reach the state shown on the right, which would be required for what I asked for.


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## Robert-Y (Apr 17, 2009)

Shoot, I was kinda getting there...


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## byu (Apr 17, 2009)

OK, I've figured out how to JUST flip the middle piece (the dot). If you want to know, read below (but you're probably not going to be satisfied with the solution).



Spoiler



Paint the yellow center white and the white center yellow.


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## Stefan (Apr 17, 2009)

Robert-Y said:


> Shoot, I was kinda getting there...


Indeed, I think you were. Well, thinking about it some more I'd now go without picture and say this:

The two stickers on the core cubie and the four outside stickers on the edge cubies always stay fixed relative to each other, no matter what moves you make. Just like the six center stickers on a 3x3x3. So having the two core stickers switch places while keeping the outer four stickers where they are is not possible.

Sorry I obsess so much over this, but I really had trouble proving it clearly (and I wasn't alone) and I'm happy I now found something I'm satisfied with. I'm particularly happy I didn't even have to talk about the corners at all.

Well, I do have to use the corners to prove that the double spot pattern isn't possible. Wait, didn't I just prove that above? No. My above proof is for the original request to "just flip the core upside down" and keep the rest of the puzzle exactly the same. But... that doesn't prove I can't build the double spot pattern a different way, by changing the outer ring so it's different but still solved. The way to disprove *that* is to observe that the outer ring can only be built one way, because each piece has forced neighbours.


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## Robert-Y (Apr 19, 2009)

"The two stickers on the core cubie and the four outside stickers on the edge cubies always stay fixed relative to each other, no matter what moves you make. Just like the six center stickers on a 3x3x3. So having the two core stickers switch places while keeping the outer four stickers where they are is not possible."

I'm giving this a thumbs up


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## Tim Reynolds (Apr 19, 2009)

cpt.Justice said:


> Tim Reynolds said:
> 
> 
> > 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!
> ...



Usually the conditions of a while loop just get checked at the end of the loop, not after every step within the loop, and you never reach that end.



cpt.Justice said:


> Tim Reynolds said:
> 
> 
> > Step 3 is confusing...I don't know which option to choose...
> ...


Right, but an algorithm should tell me which to do, since I'm lazy and indecisive.


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## IAmAPerson (May 29, 2015)

A bit late (only about 6 years), but oh well. I want to tackle the amount of algorithms needed if I were to solve the puzzle at once. It probably wasn't worth it, but I went ahead and memorized the solution for every single floppy cube scramble (oh well).

I have these on my "website" (here). To solve the 3x3 faces of the puzzle in one algorithm (assuming you know how to mirror algorithms and perform x, y, and z rotations), you need about 22 algorithms. Assuming we take a solved Floppy cube and reverse those algorithms, we should arrive at these cases, yes? Now let's perform the algorithm _R U M D L_ to swap around the edges. That's +1 case. Now let's reverse the face-solving algorithms on our new cube. That's 22 more cases. We have now a total of 45 algorithms you'd need to learn (assuming one and two-move solutions are still algorithms). I have a demonstration of the "algorithmic advanced solution" here. Note that the second solve was intuitive. I couldn't recall the algorithm, so in a panic, I simply blindsolved it intuitively. That should explain the definitely not-optimal solution.


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