Numbering the insides of a Rubiks Cube uniquely
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Numbering the insides of a Rubiks Cube uniquely
This is a puzzle I just came up with, but it is also a question I hope to get the answer to, as it's needed for a project I'm developing.
Suppose I have a Rubiks Cube in some random configuration. Now I take it completely apart, so I end up with 27 cubelets, each of which have 03 sides that are coloured, and the rest are black. Now I decide to put a white sticker over all the colours, so that each cubelet only contains white and black. But before I do that, I want to be able to assemble the Rubiks Cube in the exact same configuration, but without having to look at the colours, so I decide to write numbers on all the black surfaces.
Two surfaces that meet, get the same number written on both surfaces. So a corner cube could for instance have the numbers 1, 2 and 3 written on each of its surfaces. I want to number the surfaces such that I can only assemble them into one possible configuration, given the rule that only two black surfaces with the same number can meet. Since there are 27x654=108 hidden surfaces on a Rubiks Cube, I could use the numbers 1 through 54.
But what I'd like to know is the minimum number of numbers I could write on the surfaces, and still only assemble it in the correct configuration. I can obviously still see which cubelets are corner, edge and centre cubelets because of the white stickers, but along with the numbers on the black surfaces, those are the only information I get to reassemble the cube.
This probably isn't important, but just for good measure I'll point out that for my example the configuration could just as easily be an invalid Rubiks Cube. The point is not the Rubiks moves, but just having a way to identify each side as one of six possible. Thus a valid Rubiks cube with one cube twisted, giving a parity of 0 would be just as fine for my puzzle. Also keep in mind that reassembling the cube back but with one of the cubes twisted compared to the original configuration yields a different configuration in this case.
Suppose I have a Rubiks Cube in some random configuration. Now I take it completely apart, so I end up with 27 cubelets, each of which have 03 sides that are coloured, and the rest are black. Now I decide to put a white sticker over all the colours, so that each cubelet only contains white and black. But before I do that, I want to be able to assemble the Rubiks Cube in the exact same configuration, but without having to look at the colours, so I decide to write numbers on all the black surfaces.
Two surfaces that meet, get the same number written on both surfaces. So a corner cube could for instance have the numbers 1, 2 and 3 written on each of its surfaces. I want to number the surfaces such that I can only assemble them into one possible configuration, given the rule that only two black surfaces with the same number can meet. Since there are 27x654=108 hidden surfaces on a Rubiks Cube, I could use the numbers 1 through 54.
But what I'd like to know is the minimum number of numbers I could write on the surfaces, and still only assemble it in the correct configuration. I can obviously still see which cubelets are corner, edge and centre cubelets because of the white stickers, but along with the numbers on the black surfaces, those are the only information I get to reassemble the cube.
This probably isn't important, but just for good measure I'll point out that for my example the configuration could just as easily be an invalid Rubiks Cube. The point is not the Rubiks moves, but just having a way to identify each side as one of six possible. Thus a valid Rubiks cube with one cube twisted, giving a parity of 0 would be just as fine for my puzzle. Also keep in mind that reassembling the cube back but with one of the cubes twisted compared to the original configuration yields a different configuration in this case.
I have traveled from 1979 to be a member of the unofficial board Council of Elders. Phear M3
Re: Numbering the insides of a Rubiks Cube uniquely
4 numbers is enough to lock down the configuration of a slice of a cube (4 numbers around the center piece, then the corners of the slice have the numbers the that their two adjacent pieces lock to the center with) and 2 more to lock the slices together gives us an upper bound of 14 numbers.
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Re: Numbering the insides of a Rubiks Cube uniquely
Spoilered in case people want to think for themselves first:
Ninja'd by WarDaft's post. I think that if you combine both strategies, you can get a smaller number.
Spoiler:
Ninja'd by WarDaft's post. I think that if you combine both strategies, you can get a smaller number.
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Re: Numbering the insides of a Rubiks Cube uniquely
WarDaft, I see what you mean, but putting 4 numbers on 4 surfaces of the center piece would ensure that the 4 edge cubes were placed correctly. But then I could swap the corner cubes of that slice around anyway I wanted to. Or am I misunderstanding you?
Also, I forgot to state that the same number can be used for more than one pair of surfaces, as long as only one possible configuration can be assembled from the numbers. And even if the same number is used on two or more pairs of surfaces, it still only counts as one number.
Edit:
I realized now, upon reading Jplus' post, that I forgot to specify one more thing. The configuration only refers to the white/coloured surfaces, not to the black ones. Thus the center cube can be turned ever which way and it will be part of the same configuration either way. So really, it wouldn't be necessary to put numbers on its surfaces or those surfaces that it touches. This is because the centre cube is unique because there's only one of it.
Also, I forgot to state that the same number can be used for more than one pair of surfaces, as long as only one possible configuration can be assembled from the numbers. And even if the same number is used on two or more pairs of surfaces, it still only counts as one number.
Edit:
I realized now, upon reading Jplus' post, that I forgot to specify one more thing. The configuration only refers to the white/coloured surfaces, not to the black ones. Thus the center cube can be turned ever which way and it will be part of the same configuration either way. So really, it wouldn't be necessary to put numbers on its surfaces or those surfaces that it touches. This is because the centre cube is unique because there's only one of it.
I have traveled from 1979 to be a member of the unofficial board Council of Elders. Phear M3
Re: Numbering the insides of a Rubiks Cube uniquely
Consider:
Does this not satisfy your placing criteria?
Code: Select all
X1 1X3 3X
2 2 2
2 2 2
X1 1X3 3X
4 4 4
4 4 4
X1 1X3 3X
Does this not satisfy your placing criteria?
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Re: Numbering the insides of a Rubiks Cube uniquely
Hmm, actually... if we use a few more numbers to glue the slices together, we can reuse 1,2,3 and 4 in each of them. That brings it down to 8.
All Shadow priest spells that deal Fire damage now appear green.
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Re: Numbering the insides of a Rubiks Cube uniquely
I can do this with 6 numbers. Simply assign a number to each of the cutting plains of a rubik's cube, and put that number on every black sticker that touches it.
Re: Numbering the insides of a Rubiks Cube uniquely
WarDaft wrote:Consider:Code: Select all
X1 1X3 3X
2 2 2
2 2 2
X1 1X3 3X
4 4 4
4 4 4
X1 1X3 3X
Does this not satisfy your placing criteria?
Using the same pieces, you could also get this :
Code: Select all
X1 1X3 3X
2 2 4
2 2 4
X1 1X3 3X
4 4 2
4 4 2
X1 1X3 3X
Re: Numbering the insides of a Rubiks Cube uniquely
Spoiler:
My guess, based on intuition alone, is that the answer is
Spoiler:
Re: Numbering the insides of a Rubiks Cube uniquely
Yat wrote:Using the same pieces, you could also get this :Code: Select all
X1 1X3 3X
2 2 4
2 2 4
X1 1X3 3X
4 4 2
4 4 2
X1 1X3 3X
Hmm, that does satisfy the numbering, but conveniently, gluing the layers together actually solves that problem (as we cannot then flip any of the pieces)
All Shadow priest spells that deal Fire damage now appear green.
Big freaky cereal boxes of death.
Re: Numbering the insides of a Rubiks Cube uniquely
I'm happy to see that there is being made so much effort on solving this, and I thank you for the guesses you have made so far. I need to try the suggestions out physically because it's hard for me to wrap my mind around how they will work. But I'll point one thing out that I have realized reading the thread. Not all surfaces need a number. Take for instance a Rubiks Cube from which a corner cubelet is then detached. Only one of the cubelet's surfaces would then need a number because if the face it was touching before, also had that same number, there would only be one possible way of attaching the cubelet again.
I don't know if this will actually have any effect on the minimum number, but I thought I should point it out for good measure.
I don't know if this will actually have any effect on the minimum number, but I thought I should point it out for good measure.
I have traveled from 1979 to be a member of the unofficial board Council of Elders. Phear M3
Re: Numbering the insides of a Rubiks Cube uniquely
I think if you remove the restriction that you have to number all sides, then that reduces the numbers needed by one, since you can just count 'blank' as one of your number pairs. So I think that would give:
Spoiler:
Re: Numbering the insides of a Rubiks Cube uniquely
I was wondering, how can you tell for sure if your suggestion yields one or more solutions? There have been some good suggestions, but I simply cannot figure out how many solutions it has, without having to try out every combination. Xias made a great discovery, but says that it yields multiple configurations, which I'd like to try and understand why that's the case.
But reading his post struck me with inspiration, and I'm now hoping that it can be done with just 4 numbers, with exactly one 4 on each centre and edge cubelet and none on the corners. And here's why.
But reading his post struck me with inspiration, and I'm now hoping that it can be done with just 4 numbers, with exactly one 4 on each centre and edge cubelet and none on the corners. And here's why.
Spoiler:
I have traveled from 1979 to be a member of the unofficial board Council of Elders. Phear M3
Re: Numbering the insides of a Rubiks Cube uniquely
I think the line "Two surfaces that meet, get the same number written on both surfaces." meens you cant leave a side blank but if you think you can then take 1 from my numbers.
upper and lower bounds so far:
If I get time I might write a computer program to search this as I'm intressed in the anser.
upper and lower bounds so far:
Spoiler:
If I get time I might write a computer program to search this as I'm intressed in the anser.
Re: Numbering the insides of a Rubiks Cube uniquely
ttnarg, well I could've rephrased that line differently, I guess. I wrote that to specify that two surfaces get the same number and not two different numbers. But two adjoining surfaces don't need a number. They either get the same number, or stay blank. So either way they would be exactly the same.
I was thinking of writing a program myself, but that's including the added restraints from my previous post. If you make a program for the minimum number, we would get both answers, unless of course I actually find a working configuration for mine, which would mean that 3 is in fact the lowest number. And I'm glad that my problem has gained some interest with people.
I was thinking of writing a program myself, but that's including the added restraints from my previous post. If you make a program for the minimum number, we would get both answers, unless of course I actually find a working configuration for mine, which would mean that 3 is in fact the lowest number. And I'm glad that my problem has gained some interest with people.
I have traveled from 1979 to be a member of the unofficial board Council of Elders. Phear M3
Re: Numbering the insides of a Rubiks Cube uniquely
ThemePark wrote:I was wondering, how can you tell for sure if your suggestion yields one or more solutions?
Good question.
My original solution basically says that it's possible to uniquely solve a Rubik's cube that has been deconstructed in the way you describe using only three numbers (or two and a blank but I think the difference is unimportant). That is, it's possible to uniquely identify and rotate (in the same way that putting colored stickers on the outside of each piece does) each piece with three numbers. Someone with knowledge of which position each piece has could use that unique identification to solve the puzzle.
However, a person who does not know which position each piece has (ie someone to whom you hand all of the pieces to) might not be able to solve it, which is why I included the option for multiple configurations. I actually have no idea if multiple configurations are possible. It might be that they aren't, and any attempt to put the wrong piece somewhere will lead to a contradiction later on a la Sudoku. I don't know and don't have the capacity to find out.
So, finding a method to prove whether there are multiple solutions could prove the optimal numbers needed to be three (two with a blank).
EDIT: This puzzle reminds me of those little cardboard puzzles where you have a bunch of squares with half of a picture on each edge, and you have to arrange them so that each edge matches up.
Re: Numbering the insides of a Rubiks Cube uniquely
Xias, let me see if I understood you correctly. You mean that anyone who knew that cubes with 3 black surfaces would be corner cubes, cubes with 4 edge cubes and cubes with 5 centre cubes could solve it, while someone who wouldn't know this could find multiple configurations?
I have traveled from 1979 to be a member of the unofficial board Council of Elders. Phear M3
Re: Numbering the insides of a Rubiks Cube uniquely
ThemePark wrote:Xias, let me see if I understood you correctly. You mean that anyone who knew that cubes with 3 black surfaces would be corner cubes, cubes with 4 edge cubes and cubes with 5 centre cubes could solve it, while someone who wouldn't know this could find multiple configurations?
Oh, no, of course not. In fact, I've been picturing an actual deconstructed rubik's cube, which has clearly defined corners, edges, and centers (because each piece is rounded out in a way that allows the toy to hold together while being twisted).
What I meant was, say you have four triangles with this setup. One of them is labeled with a 1 on each edge, one labeled 123, one 134, and one 142. This allows for two possible unique orientations (Try it). However, say you know that the correct solution has each number adjacent to it's twin.s Then you can arrange them correctly without any prior knowledge.
Essentially, if the rule is to label the pieces in such a way that they can be logically solved following only the rule "Each face must touch a face with the same label", then my 3number solution would require a proof that there are no extra configurations. If having a list that says, for instance that the corner piece labeled 112 is the whiteblueorange corner piece, and so on for all of the pieces, is allowed, then my 3number solution is optimal (because you only need 3 numbers to uniquely identify and rotate each piece, meaning you cannot confuse any two different pieces or the same piece in a different orientation.)
Re: Numbering the insides of a Rubiks Cube uniquely
@ttnarg: Do you have any ideas on how to check whether there is only one possible configuration with a given distribution of numbers? All I can think of is having to try every single combination of cubelets, save rotation of course, or to start off with one layer and then adding one more if the first is valid. But both ways seem to me like they'd take a lot of time to comb through.
@Xias: I am honestly still not sure I understand what you mean. If I got it right, you're talking about only getting the information and then having to make the triangles yourself, which would lead to these two solutions.
which depends on how you write the numbers on the 123 piece. If that is the case, I intend on writing the numbers on the cubes for my project, instead of just giving the numbers. If I got you wrong, then please let me know, I'd really like to understand this. Also, I'm not sure what you mean by each number being adjacent to its twin, which is the requirement for the puzzle I stated anyway.
@Everybody: I got a lot of inspiration from Xias' post at January 16th @6:34pm UTC. So since then I have been mixing the requirements for my own project with the puzzle I originally stated, as I wanted to find the solution for this puzzle in order to further work on the project. But I have decided to go another way after Xias' post. So because I am now also interested in finding the minimum number, and to avoid any confusion, I'll restate the original puzzle that this thread is for solving, and then I'll try and figure out the puzzle for my project myself, based on the information I gathered here, which I thank all of you for.
The idea is to take a 3x3x3 cube that has some way of obtaining unique configurations. I chose a Rubik's Cube, but it could be any 3x3x3 cube for that matter. And the parity issue of the Rubik's Cube is not important either, the only point is having a 3x3x3 cube with a known and unique configuration. And also, using the Rubik's Cube, the mechanics aren't important, for the purpose of this puzzle just think of it as 27 cubes that are the exact same size and shape.
So I have such a cube in my hand. I remember the exact configuration it was in, and I then put white stickers on all 54 visible faces so that by looking at it, it's not possible to figure out the configuration, you'd actually have to peel off some stickers but that's not allowed. Having done that, I then disassemble the cube into its 27 cubelets. All 54 faces that were hidden before on the inside are black, and I write a number, a letter or a symbol on them. It doesn't matter what it is, but let's go with numbers for this. Each black face HAS to have a number, and if two black faces met before the disassembly they get the same number. Thus it can be done with 27 pairs of numbers.
The idea is then to number the black face pairs in such a way that I can give the 27 cubelets to any person, even someone who doesn't know about Rubik's Cubes, and tell that person that their job is to assemble the cube back together, by pairing off the black faces with the same number on them and ending up with a 3x3x3 cube. When that person has done that, I then peel off the white stickers to ensure that he has done so correctly.
The idea of the puzzle is to find the minimum number of numbers/letters/symbols that can be applied to all black surfaces, but in such a way that it's only possible to reach ONE configuration. There might be 10 1s which is fine, as long as all but one way of pairing off those 1s and the rest of the numbers are invalid. If there's more than one way, this way of placing the numbers is not a valid solution.
And that's the description of the puzzle. The reason I mentioned blanks at one point is because I didn't want to put numbers on all black pairs. But as it has pointed out, the minimum number is just one less if you use blanks, as one of the numbers used in the solution can be replaced by a blank surface. So blanks are only important to my project.
@Xias: I am honestly still not sure I understand what you mean. If I got it right, you're talking about only getting the information and then having to make the triangles yourself, which would lead to these two solutions.
Code: Select all
________ ________
\ 3 ^ 3 /
\ / \ /
\2 1/1 1\1 4/
\ / \ /
V___1___V
\ 1 /
\2 4/
\ /
V
________ ________
\ 2 ^ 2 /
\ / \ /
\3 1/1 1\1 4/
\ / \ /
V___1___V
\ 1 /
\3 4/
\ /
V
which depends on how you write the numbers on the 123 piece. If that is the case, I intend on writing the numbers on the cubes for my project, instead of just giving the numbers. If I got you wrong, then please let me know, I'd really like to understand this. Also, I'm not sure what you mean by each number being adjacent to its twin, which is the requirement for the puzzle I stated anyway.
@Everybody: I got a lot of inspiration from Xias' post at January 16th @6:34pm UTC. So since then I have been mixing the requirements for my own project with the puzzle I originally stated, as I wanted to find the solution for this puzzle in order to further work on the project. But I have decided to go another way after Xias' post. So because I am now also interested in finding the minimum number, and to avoid any confusion, I'll restate the original puzzle that this thread is for solving, and then I'll try and figure out the puzzle for my project myself, based on the information I gathered here, which I thank all of you for.
The idea is to take a 3x3x3 cube that has some way of obtaining unique configurations. I chose a Rubik's Cube, but it could be any 3x3x3 cube for that matter. And the parity issue of the Rubik's Cube is not important either, the only point is having a 3x3x3 cube with a known and unique configuration. And also, using the Rubik's Cube, the mechanics aren't important, for the purpose of this puzzle just think of it as 27 cubes that are the exact same size and shape.
So I have such a cube in my hand. I remember the exact configuration it was in, and I then put white stickers on all 54 visible faces so that by looking at it, it's not possible to figure out the configuration, you'd actually have to peel off some stickers but that's not allowed. Having done that, I then disassemble the cube into its 27 cubelets. All 54 faces that were hidden before on the inside are black, and I write a number, a letter or a symbol on them. It doesn't matter what it is, but let's go with numbers for this. Each black face HAS to have a number, and if two black faces met before the disassembly they get the same number. Thus it can be done with 27 pairs of numbers.
The idea is then to number the black face pairs in such a way that I can give the 27 cubelets to any person, even someone who doesn't know about Rubik's Cubes, and tell that person that their job is to assemble the cube back together, by pairing off the black faces with the same number on them and ending up with a 3x3x3 cube. When that person has done that, I then peel off the white stickers to ensure that he has done so correctly.
The idea of the puzzle is to find the minimum number of numbers/letters/symbols that can be applied to all black surfaces, but in such a way that it's only possible to reach ONE configuration. There might be 10 1s which is fine, as long as all but one way of pairing off those 1s and the rest of the numbers are invalid. If there's more than one way, this way of placing the numbers is not a valid solution.
And that's the description of the puzzle. The reason I mentioned blanks at one point is because I didn't want to put numbers on all black pairs. But as it has pointed out, the minimum number is just one less if you use blanks, as one of the numbers used in the solution can be replaced by a blank surface. So blanks are only important to my project.
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Re: Numbering the insides of a Rubiks Cube uniquely
In theory, if you are allowed to make your numbers as high as you want, then you only need one number, and you use godel numbering to encode the right configuration.
I have discovered a truly marvelous proof of this, which this margin is too narrow to contain.
Re: Numbering the insides of a Rubiks Cube uniquely
tomtom2357 wrote:In theory, if you are allowed to make your numbers as high as you want, then you only need one number, and you use godel numbering to encode the right configuration.
Except that you can't distinguish between two corner pieces if you're only using one number. It doesn't matter if you encode the configuration for a cube, all 8 corners will have the same encoding, and you will have no way to tell which goes where.
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Re: Numbering the insides of a Rubiks Cube uniquely
Okay then, you use 7 numbers. The corner closest to you uses the godel numbering as before, then you proceed to go in a clockwise direction around the three other close corners (incrementing by 1 each corner so you can tell the difference), then you do the same for the other three corners farther away, and you leave the opposite corner blank, and you have effectively distinguished between the corners. You could probably do less, you just need to break the symmetry, the godel numbering encodes the rest.
I have discovered a truly marvelous proof of this, which this margin is too narrow to contain.
Re: Numbering the insides of a Rubiks Cube uniquely
Okay, so after having given this some thought, I have a gut feeling that I have come up with a way to numbering the black sides using only 3 sides, and doing so uniquely. But I haven't done any testing on this, so feel free to prove me wrong.
To apply the numbers, I start out with a Rubiks Cube in its solved state. I then replace all the colours with a number from 1 to 6, both inclusive, and I do this in such a way that the rules of a dice apply, i.e. two opposite sides add up to 7. Now I take the cube apart, making sure to remember where each cubelet goes, and I add the remaining numbers on the black sides of the corner, edge and middle cubes, again using the dice rules. That leaves the centre dice which can't initially be numbered since only one side is known, but using the numbers of the sides touching it, they too can be numbered, that's why you need to remember their position up until this point. Finally, to go from 6 to 3 numbers, I simply use modulo 3 on all numbers, so 1=1, 2=2, 3=0, 4=1, 5=2, 6=0.
And I believe this way of numbering the cubelets should ensure a unique configuration once you reassemble them. But I'm hoping someone will be able to try this out and prove me right or wrong.
To apply the numbers, I start out with a Rubiks Cube in its solved state. I then replace all the colours with a number from 1 to 6, both inclusive, and I do this in such a way that the rules of a dice apply, i.e. two opposite sides add up to 7. Now I take the cube apart, making sure to remember where each cubelet goes, and I add the remaining numbers on the black sides of the corner, edge and middle cubes, again using the dice rules. That leaves the centre dice which can't initially be numbered since only one side is known, but using the numbers of the sides touching it, they too can be numbered, that's why you need to remember their position up until this point. Finally, to go from 6 to 3 numbers, I simply use modulo 3 on all numbers, so 1=1, 2=2, 3=0, 4=1, 5=2, 6=0.
And I believe this way of numbering the cubelets should ensure a unique configuration once you reassemble them. But I'm hoping someone will be able to try this out and prove me right or wrong.
I have traveled from 1979 to be a member of the unofficial board Council of Elders. Phear M3
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