Storage in nD, how compact it can be?

[Secret]

Today someone from fb shared me this
http://www.youtube.com/watch?v=dAa6bOWB ... e=youtu.be
And I am amazed on how this guy can make use of space so effectively

Now I am thinking, since in 4D we have essentially a whole plane (hedrix) of sideways directions to store things, so if we scale up e.g. the compact bed thing in the video to 4D, how small it can look when completely folded up

This also caused me to revisit one of alkaline's intro pages and compare how in 2D it will work
Is it possible to have as efficient storage space in 3D in a 2Der's world?

One a more mathematical context: Is there any formula that can illustrate the trend in the efficiency on storage space in different number of spatial dimensions?

NB: This topic I raised in a kinda casual tone ,because I am currently focusing on the toratopes

[quickfur]

I didn't watch the video (I'm at work, and office PC doesn't have Java/Flash installed), but generally, assuming you have a room in the shape of an n-cube (or n-cuboid), then the higher the dimension, the more space you get at the corners of the n-cube.

To get some idea about this, imagine if you have an n-cube shaped room and you need to pack as much stuff as you can inside, while still leaving enough room to walk through the middle of the room. The simplest model of this is to cut the floor space of the n-cube, which is in the shape of an (n-1)-cube, into 3x3x3x...x3 smaller (n-1)-cubes, like an (n-1)-dimensional Rubik's cube, and then designate the cubies that lie along the coordinate axes to be floor space reserved for walking through the room. So this leaves 2^(n-1) cubies at the corners of the (n-1)-cube shaped floor space available for furniture and storage boxes. The floor space is limited to 2(n-1) cubies that touch the walls of the room, plus 1 cubie in the middle that connects all of the walking space cubies, which makes 2(n-1)+1 cubies of floor space.

Now notice that the corner cubies, which are to be used for storage, are exponential in number, whereas the number of unoccupied floor space cubies are only linear in number. So this means that the higher the dimension you go, the more storage space there will be, given a room with the same edge lengths. Exponentially more floor space, in fact. Every additional dimension you go doubles the amount of cubies dedicated for storage, while only adding 2 more cubies of walking room. In 3D, you have a 3x3 floor grid, and 5 squares reserved for walking, and 4 for storage: rather inefficient. But in 4D, while the number of cubies reserved for walking increases to 7, the number of storage cubies doubles to 16, plus 12 newly-introduced edge cubies that are also available for storage. So that's 7 walking spaces to 28 storage spaces. Already, 4 times floor space is dedicated for storage than for walking. Going up to 5D, the number of walking spaces is 9, but the number of storage corners doubles again to 32, but now there's a whole lot more edge spaces that can also be used for storage, bringing the total to 72 storage spaces. Now there is way more storage space than walking space, and yet you can reach any of the 72 storage areas just by standing in the center walking space. In 6D, the number of walking spaces is 11, but the number of storage areas is 232. The ratio of storage spaces to walking spaces is rapidly increasing. In 7D, there are 13 walking spaces and a whopping 716 storage spaces. In 8D, there are 15 walking spaces and 2172 storage spaces. By the time you get to 10D, there's 19 walking spaces and 19,664 storage spaces: the storage-to-empty-space ratio is now more than 1000. Go up to 100D, and you have 198 walking spaces, and an insane number of storage spaces: 171,792,506,910,670,443,678,820,376,588,540,424,234,035,840,468 (that's 1.7*10^47 storage spaces!!). By this point, basically the entire room is storage space, and yet you have ample space to walk through without any obstruction!

Now, the killer is that the above calculations are actually based on a rather poor packing of things in an n-cube. There are far more efficient packings possible, for example, the Leech lattice packing in 24 dimensions where you can fit 16,777,216 24-spheres around a single 24-sphere, which can then be enclosed by a box of the right size. Even after you reserve some number of spheres as "walking space", there's still an incredible amount of available packing space there.

So yeah, one of the effects of going up to higher dimensions is that there is a lot more available space, and you can squeeze a lot of stuff into what we 3Ders would consider a small space (but in actuality it's not a small space at all, it's just that higher dimensional space is just sooooo spacious, if you're excuse the pun ).

[Klitzing]

Hi quickfur, I never could get some rhime out of that proposal of yours. Now, as you outlined your underlying idea I can come in. You just compares the count of corners to that of facets plus one (center). That seems somehow incorrect. I'd think you should rather compare V : (E + F + C + ...) here.

Applied to the series of hypercubes this then runs like

Code: Select all

nD | V : (E + F + C + ...)
---+----------------------------
1D |  2 : 1  = 2.0
2D |  4 : (4+1)  = 0.8
3D |  8 : (12+6+1)  = 0.421
4D | 16 : (32+24+8+1)  = 0.254
5D | 32 : (80+80+40+10+1)  = 0.152


One further notes that for hypercubes we have V + E + F + C + ... = 3^(n-1) and V = 2^n. Accordingly that series runs like 2^n / (3^(n-1) - 2^n). Obviously in the denominator 3^(n-1) overcomes the 2^n by far. Thus in the limit you could consider 2*(2/3)^(n-1) instead, which clearly decreases towards zero.

--- rk

[quickfur]

Hi quickfur, I never could get some rhime out of that proposal of yours. Now, as you outlined your underlying idea I can come in. You just compares the count of corners to that of facets plus one (center). That seems somehow incorrect. I'd think you should rather compare V : (E + F + C + ...) here.

Applied to the series of hypercubes this then runs like

Code: Select all

nD | V : (E + F + C + ...)
---+----------------------------
1D |  2 : 1  = 2.0
2D |  4 : (4+1)  = 0.8
3D |  8 : (12+6+1)  = 0.421
4D | 16 : (32+24+8+1)  = 0.254
5D | 32 : (80+80+40+10+1)  = 0.152


One further notes that for hypercubes we have V + E + F + C + ... = 3^(n-1) and V = 2^n. Accordingly that series runs like 2^n / (3^(n-1) - 2^n). Obviously in the denominator 3^(n-1) overcomes the 2^n by far. Thus in the limit you could consider 2*(2/3)^(n-1) instead, which clearly decreases towards zero.

--- rk

Well, there are two ways to think about this. One is that you only store stuff at the vertices of the n-cube, as you describe, in which case most of the room will be empty space, much of which is probably unnecessary.

But what I have in mind is that you reserve 2n+1 cubies from the n-cube as empty space, that is, exactly those that lie on the coordinate axes (the last of which lies on the origin). I think this makes more sense because if you're using the room for storage, you'd want to maximize storage space, not minimize it to only the corners! The 2n+1 n-cubes of reserved space is to permit easy access to all storage areas. When you're standing on the center n-cube, you can reach all other cubies because they all share at least 1 vertex with that center, so this is already enough to permit easy access to all the storage areas.

Of course, you can also do something in-between, in which case you'll get varying ratios of space to storage. The point, though, is that it is possible to maximize storage area without compromising easy access, and the efficiency increases without bound as the dimension increases.

[Keiji]

I don't think people would store stuff as you have described, quickfur - you would need to share an (n-1)-dimensional facet with the cubie you are storing things in in order to reach it, surely?

[quickfur]

I don't think people would store stuff as you have described, quickfur - you would need to share an (n-1)-dimensional facet with the cubie you are storing things in in order to reach it, surely?

Well, that analysis was just for dividing up the floor space. Presumeably, you won't be stacking stuff to the ceiling, so there should still be an (n-1)-dimensional facet on the top of the storage spaces where you can access it.

You do have a good point, though, that things are harder to access if you have to reach diagonally across other storage boxes in order to reach it, even if the top is accessible.