solving with missing stickers?

[rjohnson_8ball]

What is the maximum number of stickers you can remove from a 3x3 cube and still keep it fully functional (43*10^18 distinct states)? Assume the cube is assembled correctly before removing the stickers and scrambling it.

Spoiler

I think 21 stickers can be removed, if done properly.

 

[Cride5]

21?

Reasoning:

Spoiler

Centre Stickers: 4 - Assuming a particular colour scheme, only 2 adjacent centre stickers are required to uniquely identify cube orientation.

Corner Stickers: 10 - Remove 1 from each corner, since they can be uniquely identified by only two stickers. No corners may have all stickers removed since its impossible to determine orientation if this is the case*. Of the 8 corners, only 2 (each in opposite layers) can have 2 stickers removed (namely those stickers not belonging to the opposite faces) since its possible to uniquely determine the fourth corner in each layer by elimination of the other 3.

Edge Stickers: 6 - Its not possible to remove 2 stickers from any edges since it is impossible to determine orientation*. Among the edges with one sticker removed, the remaining colour must be unique. This means a maximum of 6 stickers may be removed from the edges, leaving 6 edges with exactly one sticker.

*Finally: One sticker can be removed from either an edge or corner with one remaining sticker (despite the earlier contradictory statements), since its orientation and permutation is determined by the orientation/permutation of all other pieces.

4 + 10 + 6 + 1 = 21

Solving a cube like this would be quite an interesting challenge actually..

 

[SkateboardingCuber]

This is very interesting. Nice reasoning, guys

 

[Thomas09]

21?

We all know what that's half of.

 

[Lucas Garron]

Hmm, I don't know why my post never showed up. Anyhow, I got to the same conclusion as Cride.

 

[qqwref]

21?

Reasoning:

Spoiler

Centre Stickers: 4 - Assuming a particular colour scheme, only 2 adjacent centre stickers are required to uniquely identify cube orientation.

Corner Stickers: 10 - Remove 1 from each corner, since they can be uniquely identified by only two stickers. No corners may have all stickers removed since its impossible to determine orientation if this is the case*. Of the 8 corners, only 2 (each in opposite layers) can have 2 stickers removed (namely those stickers not belonging to the opposite faces) since its possible to uniquely determine the fourth corner in each layer by elimination of the other 3.

Edge Stickers: 6 - Its not possible to remove 2 stickers from any edges since it is impossible to determine orientation*. Among the edges with one sticker removed, the remaining colour must be unique. This means a maximum of 6 stickers may be removed from the edges, leaving 6 edges with exactly one sticker.

*Finally: One sticker can be removed from either an edge or corner with one remaining sticker (despite the earlier contradictory statements), since its orientation and permutation is determined by the orientation/permutation of all other pieces.

4 + 10 + 6 + 1 = 21

Solving a cube like this would be quite an interesting challenge actually..

I think I can get 24.

Spoiler

Centers - 4 stickers is minimum.

Corners - I think 8+5 can be removed without parity considerations. Imagine that the corners belonging in URF, URB, ULF, DRF are known. Then, by process of elimination, as long as the correct single sticker is on the three adjacent corners (ULB, DLF, DRB), we can use process of elimination to prove exactly which corner each one is. The last corner doesn't need any stickers because its position and orientation are determined by the other corners.

Edges - Counting the "parity sticker" I think 7 is indeed optimal. I agree that the remaining color must be unique (although we can have one duplicate due to parity) so there's no way we can do better than 7.

4+13+7 = 24.

PS, Lucas: You removed 6 edge stickers, but you allow edge parity; however you have removed 11 corner stickers and corner permutation is determined. This should be a signal that Cride's analysis is not quite optimal.


PPS: Here's another interesting question. Is there a partially stickered cube which is an optimal-least-stickers-unique-solution cube for TWO different color schemes?


PPPS: Here's a picture of 24 removed stickers, which again I think is the highest number you can remove and still get a unique solution. I dare you to sub20 a cube like this.

 

[Cride5]

Spoiler

Corners - I think 8+5 can be removed without parity considerations. Imagine that the corners belonging in URF, URB, ULF, DRF are known. Then, by process of elimination, as long as the correct single sticker is on the three adjacent corners (ULB, DLF, DRB), we can use process of elimination to prove exactly which corner each one is.

Good point! Instead of taking one from two opposite layers its possible to take from three adjacent layers. Nice observation.

I realised I made another error with the orientations. Instead of taking all stickers from an edge OR a corner, its possible to take all stickers from a single edge AND a single corner since all other edges/corners determine orientation in both cases. So in my proof I should really have counted 22. Bit of a cube theory fail :rolleyes:

EDIT: This theory question has an interesting speedsolving perspective. I noticed when I was competing in the UK open, although I did no full cube rotations I was wasting a lot of time 'looking' around the cube. I've since been trying to recognise my F2L (and LL) cases using only the U, F, L and R faces. Using 2 stickers to identify corners, or using a process of elimination to identify partially hidden pieces comes into it quite a bit!

 

[Lucas Garron]

Nice, qq. My re-post totally didn't convey my original doubt, and I didn't really care to take the time to do better.

If I ever dip another cube in liquid nitrogen, this means I can afford to lose 24 of those 30-year-old stickers.

PPS: Here's another interesting question. Is there a partially stickered cube which is an optimal-least-stickers-unique-solution cube for TWO different color schemes?

Isn't it not a unique solution, then? I understand you to mean "a stickering with N=2 solutions such that every 6-face coloring scheme has at most M=1 solutions with the stickering."

Anyhow, how about leaving out the E-ring? It could provide four solutions, but allows for a parity mess.

 

[rokicki]

I think I can get 24.

Hmm, this one seems to have a solution when you swap the orange and red faces, so to me this would seem *not* to have a single solution.

Or am I missing some nuance?

 

[cuBerBruce]

I think I can get 24.

Hmm, this one seems to have a solution when you swap the orange and red faces, so to me this would seem *not* to have a single solution.

Or am I missing some nuance?

The problem that was posed was to have 43252003274489856000 distinguishable states, not to have a single solved state. In this case, the two solved states are distinguishable.

 

[rokicki]

The problem that was posed was to have 43252003274489856000 distinguishable states, not to have a single solved state. In this case, the two solved states are distinguishable.

Yeah, I get that, but does multiple solutions keep it "fully functional"? Was the parenthetical comment definitive, or simply necessary but not sufficient?

 

[rjohnson_8ball]

@qqwref, I missed that. See, that's why I posed the question. I wasn't totally sure I wasn't missing something.

@rokicki, can you clarify? I don't understand.

 

[qqwref]

Hmm, this one seems to have a solution when you swap the orange and red faces, so to me this would seem *not* to have a single solution.

Or am I missing some nuance?

My interpretation of the problem was that, if you already know the color scheme, there should be exactly one possible solved position. My problem has to be interpreted in that sense - a cube that has exactly one solution (each) for two different fixed color schemes.

The problems of having exactly one solution for a given color scheme and of having ~4.3 * 10^19 distinct states seem to be the same, whereas the problem of having exactly one distinct state for which all stickers on each face are the same color is slightly more complicated (is 23 stickers the max.?).

 

[rokicki]

good clarification

Thanks for the clarification; that makes sense.

It's a good problem, and I may well sticker a cube like this just for kicks.

 

[Pembo]

Shameless self-promoting bump.

http://www.youtube.com/watch?v=qk_eMx5PRWs

44.44 PB so far.

 

[DaijoCube]

I used to solve an OLD rubik with many vanished stickers until I received new cubes and CubeSmith stickers Don't know how many were missing, but it was harder to solve I did the PseudoVoidCube stickermod to it later on.

 

[riffz]

Shameless self-promoting bump.

http://www.youtube.com/watch?v=qk_eMx5PRWs

44.44 PB so far.

Impressive. I haven't tried this yet myself but that seems like a very fast time.

 

[Pembo]

I'm hitting low40s now, it's still very disorientating.

Also, OLL is a joke, so I just solve LL in 4 step usually.

 

[chrisness]

I realised I made another error with the orientations. Instead of taking all stickers from an edge OR a corner, its possible to take all stickers from a single edge AND a single corner since all other edges/corners determine orientation in both cases. So in my proof I should really have counted 22. Bit of a cube theory fail :rolleyes:

I haven't thought this through a huge amount, but doesn't this mean we can take another sticker from QQ's solution to leave a single edge with no stickers?

 

[Pembo]

I don't see the problem with taking a green edge sticker off, would also make solving it easier.

But anyway, get back to chemistry or imma come to Clare and eat you.