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# Rubik's Hypercube: Newsletter Challenge (August 2013)

Posted July 31, 2013 5:01 PM
Pathfinder Tags: challenge questions

This month's Challenge Question: Specs & Techs from GlobalSpec:

A typical Rubik's cube is made up of 27 small cubes (3 cubes per edge). One of the cubes is always hidden. How many hidden cubes would a Rubik's cube with 5 cubes per edge have? How many hidden hypercubes would a Rubik's hypercube with 3 hypercubes per edge have (assuming both the hypercube, and the observer's perspective, are 4D)?

A Rubik's cube with 5 cubes per edge would have 27 hidden cubes. A 4D Rubik's hypercube with 3 hypercubes per edge would have 1 hidden hypercube. Both of these answers can be determined through the following formula:

In the formula above, Nh is the number of hidden cubes/hypercubes, Ne is the number of cubes/hypercubes per edge, and D is the dimension of the Rubik's cube/hypercube.

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#1

### Re: Rubik's Hypercube: Newsletter Challenge (August 2013)

07/31/2013 5:26 PM

The 5 x 5 x 5 cube is easy. 125 total cubes with 27 cubes hidden.

You have 25 cubes per side and if you count two opposing sides as left and right you have 50 exposed cubes there. Plus two sides (front back) with a 3 x 5 cubes or a subtotal of 30 cubes. Then the top and bottom is two 3 x 3 cubes or 18.

50 + 30 + 18 = 98. 125 cubes - 98 cubes = 27 cubes.

I'll do the tesseract cube count shortly...

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#2

### Re: Rubik's Hypercube: Newsletter Challenge (August 2013)

07/31/2013 5:32 PM

The professor's cube....

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#31

### Re: Rubik's Hypercube: Newsletter Challenge (August 2013)

08/08/2013 4:31 AM

I see your 5x5x5 and raise you a 9x9x9 and an 11x11x11

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#32

### Re: Rubik's Hypercube: Newsletter Challenge (August 2013)

08/08/2013 5:50 AM

That's better!

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#3

### Re: Rubik's Hypercube: Newsletter Challenge (August 2013)

07/31/2013 5:43 PM

Okay. Assuming that each tesseract cube has 7 cubes and we have 3 x 3 x 3 tesseract cubes stacked as follows:

Each side has 3 x 3 cubes or 9 cubes facing out. There are 6 sides, so 6 * 9 = 54 cubes facing out.

But we have a total of 27 tesseracts * 7 cubes each tesseract = 189 cubes.

189 total cubes - 54 cubes facing out = 135 cubes not seen, so to speak.

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#9

### Re: Rubik's Hypercube: Newsletter Challenge (August 2013)

08/01/2013 8:26 AM

A tessaract is made of 8 cubes, not 7. There is an 'extra' cube to which the 6 new 4-D cubes attach.

It's analogous to a 3-D cube formed by adding a square to each edge of a 2-D square, and then adding one more square to which these new squares attach.

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#10

### Re: Rubik's Hypercube: Newsletter Challenge (August 2013)

08/01/2013 8:28 AM

Can you provide a link for that definition?

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#12

### Re: Rubik's Hypercube: Newsletter Challenge (August 2013)

08/01/2013 8:34 AM

I give a link to the wikipedia article in my reply below (#11).

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#13

### Re: Rubik's Hypercube: Newsletter Challenge (August 2013)

08/01/2013 8:52 AM

Found it. 8 cubes. I can not yet determine how you stack such a creature together to form a larger structure (3 x 3 x 3) Rubik Hypercube.

If my illustration still technically stands, and each of the external hypercube's 8th cube is technically exposed, then the answer would be 136, as the inner hypercube is the only unexposed hypercube.

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#4

### Re: Rubik's Hypercube: Newsletter Challenge (August 2013)

07/31/2013 5:52 PM

Here is a schematic representation of a single hypercube or tesseract. All lines are at a theoretical 90Â° to each other, but it is impossible to create in 3D space, so we distort the outer cubes to represent the theoretical figure as shown:

Each 3D object represents 1 cube. You get 7 total cubes in the figure. 1 cube in the center surrounded by 6 other cubes (3D trapezoids).

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#5

### Re: Rubik's Hypercube: Newsletter Challenge (August 2013)

07/31/2013 8:24 PM

......but assuming a 4th dimension of time wouldn't all the hidden surfaces eventually expose themselves?...

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#7

### Re: Rubik's Hypercube: Newsletter Challenge (August 2013)

07/31/2013 9:17 PM

Yeah, 4D threw me off, too. I chucked it up to just a poor wording.

A hypercube is also a tesseract, and I used the standard definition of 6 cubes surrounding a central cube all folded into a larger cube.

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#8

### Re: Rubik's Hypercube: Newsletter Challenge (August 2013)

07/31/2013 11:12 PM

So it's 27 and 1....?

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#18

### Re: Rubik's Hypercube: Newsletter Challenge (August 2013)

08/02/2013 7:03 AM

It's clear that a 5x5x5 cube fully encloses a 3x3x3 cube.

For the hypercube I have no idea what all this talk of 7 and 8 is about.

There are 9 squares in a 3x3 square

There are 27 cubies in a 3x3x3 cube.

And there are 81 hypercubies in a 3x3x3x3 hyper cube. As you move (look?) up or down (forward or backward?) in any dimension the two outer elements block access to the one inner element whichever dimension you are moving in.

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#11

### Re: Rubik's Hypercube: Newsletter Challenge (August 2013)

08/01/2013 8:29 AM

You've mis-counted. There is an extra outer cube.

http://en.wikipedia.org/wiki/Tessaract

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#6

### Re: Rubik's Hypercube: Newsletter Challenge (August 2013)

07/31/2013 9:02 PM

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#14

### Re: Rubik's Hypercube: Newsletter Challenge (August 2013)

08/01/2013 9:03 AM

Since there are 8 large 'hyper-cubes' in a tessaract, each hypercube being composed of 27 small hyper-cubes, there are a total of 216 'cubes'. My guess is that a 4-D being would see 215 of the small hypercubes, with 1 always hidden.

My guess is based on an analogy to 2D and 3D Rubik cubes.

In the 2D analogy a 2D being can 'see' a total of 8 of the squares, with one square always hidden, no matter how the big square is rotated. As 3D beings, we can see all 9 squares. Here's my sketch of the 2D analog.

For us 3D beings looking at a 27-cube Rubik Cube, we can see 26 of the cubes with 1 always hidden; so analogously I think a 4D being would see all 27 of the cubes in a 3D Rubik Cube.

But a 4D being looking at a 4D Rubik cube composed of 8 large hypercubes, each composed of 27 smaller hypercube, I think he (or she) would still have 1 cube hidden, with the rest being visible. So 216 total cubes; he could see 215, with 1 hidden.

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#15

### Re: Rubik's Hypercube: Newsletter Challenge (August 2013)

08/01/2013 9:10 AM

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#16

### Re: Rubik's Hypercube: Newsletter Challenge (August 2013)

08/01/2013 9:51 AM

Thanks. I admit it's just a guess; I don't have the patience to plow through any calculations to actually do a proof of what ever the correct answer is.

So this is my answer and I'm stickin' to it.

(at least for now)

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#17

### Re: Rubik's Hypercube: Newsletter Challenge (August 2013)

08/02/2013 5:21 AM

You said: "Since there are 8 large 'hyper-cubes' in a tessaract, each hypercube being composed of 27 small hyper-cubes..." I don't get it. Why so? "Tessaract" and "Hyper-cube" are two different names for the same geometrical object (i.e. a 4D cube). A hyper-cube consists of 8 cubes (not hyper-cubes). Of course, these cubes are distorted as they are projected to our 3D space.

So again, the "4D Rubik cube" consists of 27 hyper-cubes. And, again, one of them is internal and hiden, i.e. a 4D creature is not able to see it. And, of course, a 5D creature is able to see all the 27 hyper-cubes.

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#19

### Re: Rubik's Hypercube: Newsletter Challenge (August 2013)

08/02/2013 7:20 AM

OK, I guess I am just a realist.....

There are 3x3 cubes on the top =9

There are 3x3 cubes on the bottom = 9

There are 3 cubes along 1 side = 3

There are 3 cubes along other 1 side :>) = 3

1 in between each end of the sides..... = 2

That makes 26, the center is hollow, no cube...

I know, that is how I solve the rubrics cube, I take it apart, and then re-assemble it with all the colors matching.

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#20

### Re: Rubik's Hypercube: Newsletter Challenge (August 2013)

08/02/2013 1:54 PM

Yes, tessaract and hypercube are the same thing. I was trying to distinguish between the the 'cube faces' of the hypercube and the overall hypercube -- similar to the way you start with a 2D square and add additional 'square faces' in the 3rd dimension to form a 3D cube.

Each 'face' of the '4D Rubik' is a Rubik cube (composed of 27 smaller cubes). You need a total of 8 Rubik cubes to make a 4D Rubik Cube.

A 2D 'Rubik Cube' is the 9-square pattern I sketched above (# 14).

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#24

### Re: Rubik's Hypercube: Newsletter Challenge (August 2013)

08/05/2013 4:19 AM

... So again, the "4D Rubik cube" consists of 27 hyper-cubes. And, again, one of them is internal and hiden, i.e. a 4D creature is not able to see it. And, of course, a 5D creature is able to see all the 27 hyper-cubes. ...

Good point to mention the parameter of the creature's dimension. A creature with a dimension more than that of the Rubic's (hyper-)cube will be able to see through all of its structure (all sub-cubes will be viewable from all non trivial view-points in the extra dimension), i.e. there will be no hidden (hyper-)cubes.

Only one comment here: When we talk about a 4D Rubic's cube, do we mean a 3D cube consisted of 27 smaller 4D hypercubes, or rather a 4D cube consisted of 3x3x3x3 = 81 small hypercubes? I bet it's the second. Even in this case though, there will be a 4D cube of size 1x1x1x1 which is enclosed and therefore hidden to a 4D creature.

Similarly, a Rubic's hyper-cube of size 5 will have 3x3x3x3 hidden hyper-cubes. As an extra note, a 3D creature would be unable to see the 3x3=9 of the total of 5x5=25 projected cubes, i.e. there will be 16 hidden cubes.

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#38

### Re: Rubik's Hypercube: Newsletter Challenge (August 2013)

08/19/2013 5:04 AM

Yes tkot, you are right. In fact, I considered that right after I sent my post but I hadn't the time to make a correction. A 4D Rubic cube consists of 81 small hypercubes.

You wrote: "As an extra note, a 3D creature would be unable to see the 3x3=9 of the total of 5x5=25 projected cubes, i.e. there will be 16 hidden cubes." Hmmm... ... I don't get it. A 4D Rubic cube of size 5 consists of 5X5X5X5=625 small hypercubes. So, the 3D projection should be consist of 625X8=5000cubes. Not all of them will be visible by a 3D creature but I'm not sure how many of them will be hidden.

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#21

### Re: Rubik's Hypercube: Newsletter Challenge (August 2013)

08/03/2013 8:33 PM

There aren't any cubes inside. I took one apart.

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#22

### Re: Rubik's Hypercube: Newsletter Challenge (August 2013)

08/03/2013 11:27 PM

I think some of you are way over analyzing this.

Just peal the stickers off of all of the faces and count them. That is the only way to be sure.

Plus if you do it carefully when you are done counting you can put all of them back on in the right order instead of having to figure out how to unscramble them the hard way.

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#23

### Re: Rubik's Hypercube: Newsletter Challenge (August 2013)

08/04/2013 9:11 PM

A square exists in 4 linear worlds, 2 pairs parallel worlds;

A cube exists in 6 flat worlds, 3 pairs parallel worlds;

So, a hypercube maybe exists in 8 stereoscopic worlds, 4 pairs parallel worlds.

Maybe the better way we can 'see' a hypercube is walking through the 4 pairs parallel worlds.

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#25

### Re: Rubik's Hypercube: Newsletter Challenge (August 2013)

08/05/2013 6:27 AM

Hate the so called hypercube-like projection of a n-dimensional space (n>3) cube in 3d, consider it a poor man's mind game, it only diverts from reality and as such in my opinion is next to worthless. You don't really have to visualize anything on n>3 dimension, you can ONLY apply the math that adds up from the previous (n-1) dimension in analogy to the property differences when advancind between two neigboring levels of perceivable dimensions. (i.e. 1d to 2d and 2d to 3d) So, a 4-d 5 in-line analogous to Rubic in the 4-d world thing, would consist from a total of 5^4 that is 625 objects (not cubes really eh?) from which (5-2)^4 that is 81 would be invisible from outside, by a supereyed, 4-d observer. That's as far as my 3-d hyperselfrespect will allow me to go. S.M.

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#26

### Re: Rubik's Hypercube: Newsletter Challenge (August 2013)

08/07/2013 12:54 AM

a 5 piece per line hypercube has 27 hidden cubes

the captain of the submarine is older than 5

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#30

### Re: Rubik's Hypercube: Newsletter Challenge (August 2013)

08/08/2013 12:51 AM

if the center cube o this inner cube is left, than there are 26 cubes inside the 5 by 5 by 5 cube, but this invissible inner center cube is still a (not existing) cube.

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#27

### Re: Rubik's Hypercube: Newsletter Challenge (August 2013)

08/07/2013 1:15 PM

The way I'm reading the question is a 4-D Rubik's cube with three 4-D hypercubes on each edge. It is also asking how many hidden hypercubes and not how many hidden cubes. I think this theoretical object is made of 81 hypercubes (3x3x3x3) and only one is hidden from you when you're wearing your 4-D glasses.

My logic is as follows:

1-D Universe: You have three line segments end to end in the X direction. If you can only see in the X direction (1-D vision), then ONE line segment will be hidden, and the lines you see will appear as points. This is the 1-cube (http://en.wikipedia.org/wiki/Hypercube) scenario.

2-D Universe: You have a 9 squares arranged in a 3x3 square in the XY plane. If you can only see in the XY plane (2-D vision), then only ONE square will be hidden, and the squares you see will appear as lines. This is the 2-cube.

3-D Universe: You have 27 cubes arranged in a 3x3x3 cube. If you have access to all 3 dimensions, like we typically do, only ONE cube is hidden and the cubes you see will appear as squares. This is the 3-cube.

4-D Universe: You have 81 4-D hypercubes arranged in one 3x3x3x3 4-D hypercube. I suspect that if you had 4-D vision, then only ONE hypercube would be hidden, and the hypercubes would appear as cubes. This is the 4-cube.

Having said all that, I still have no idea how to wrap my mind around a 4-D object.

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#28

### Re: Rubik's Hypercube: Newsletter Challenge (August 2013)

08/07/2013 5:28 PM

Exactly: GA.

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#33

### Re: Rubik's Hypercube: Newsletter Challenge (August 2013)

08/09/2013 12:56 AM

If you have access to all 3 dimensions, like we typically do, only ONE cube is hidden and the cubes you see will appear as squares. This is the 3-cube.

you look in direction ofr the Z-Axis onto the X-Y-Plane and you will see only 9 squares! There will be not one cube visible!

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#29

### Re: Rubik's Hypercube: Newsletter Challenge (August 2013)

08/07/2013 5:37 PM

There are a lot of strange ideas kicking round in this thread.

The normal 2d picture of a 3d representation of a 4d cube:-

is analogous to this 2d representation of a 3d cube:-

You can kind of see that it has 6 square faces albeit 4 are somewhat distorted.

We all know what a typical 2d representation of a 3d Rubik's cube looks like, but another way to represent it could be like this:-

We have to acknowledge that the three slabs have a thickness of 1, and, that when they are stacked up in the z direction 1 on top of 2 on top of 3 then they would all fall in exactly the same place in the x and y directions.

Now, the fact that it has six square faces doesn't affect our calculation of 3x3x3 cubes in a 3d cube with sides equal to 3 units. And there is no reason why the fact that a 4d cube has eight 3d cubes as faces should affect our similar 3x3x3x3 calculation.

This is a 2d picture of a 3d representation of a 4d cube which is analogous to the 2d representation above:-

Again we have to acknowledge that the blocks have a thickness of 1 in the fourth dimension w say, and that when the three blocks are stacked up, they all occupy exactly the same positions in x, y, z space. I have just highlighted the middle cube because the above picture is part of the one below.

A 4d being can move and see smoothly between blocks 1, 2 and 3 but clearly it has to stop when it hits something solid. In the x, y and z directions we all understand how it's easy to see and touch all the cubes except the middle one in block 2 but what about the middle ones in blocks 1 and 3.

If it (the 4d being) starts in the cube in block zero (there is nothing there it's just a space) it can move in the w direction until it reaches the cube in the centre in block 1, but it can't get through it to the one in the centre of block 2. Similarly block 4 to 3.

If this still doesn't make sense try following the same reasoning for the 2d representation of the 3d cube above.

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#34

### Re: Rubik's Hypercube: Newsletter Challenge (August 2013)

08/10/2013 8:34 PM

Great question, and so difficult to visualise. Trying to cram a hypercube into your head just gives you a headache. So I am with Bioengeneericist on this one.

If you take a point (0d), you have no directions from which to look as a 0d person. If you take the point, and draw it out along a line (1d) for 3 units, a 1d person can look at it from each end, but one unit (the middle one) will be hidden from veiw. If you then take that line, and draw it out along a second dimension for 3 units, you get a square, 3 "blocks" on each side. A 2d person can look at this square from 4 different directions, with the one central square obscured from view. If you then take that square, and draw it out along a third direction for 3 units, you get a cube, 3x3x3 blocks, which can be veiwed by us 3d mortals from 6 different directions, again with the one central cube obscured from view.

If you then take that cube, and draw it out along a fourth dimension for 3 units (you can imagine this visually only by analogy in 3d), you get a hypercube, 3x3x3x3 hyper blocks, which can be viewed from 8 different directions by a 4d person. Again, one central hyperblock will be obscured from veiw.

So my guess, concurring with others, is that the answer is 1.

If you think about this a bit more, the 1d person sees the faces of their cube as points (2 points from 2 directions), the 2d person sees the faces as lines (4 lines from 4 directions), the 3d person sees square faces on their cubes (6 square faces from 6 directions), and the 4d person sees cubic faces on their hypercube (8 cubes each seen as a "surface", from the 8 directions). Just as we can see all the blocks on the face of a rubic cube in 3d, a 4d person would be able to see all 27 cubes making up one face of their hypercube, even the central one. So none of these 8 central cubes are hidden. Only the one at the core of the hypercube.

Just as a 3d person looking at a 3x3 square (the 2d Rubic cube) can see all 9 blocks, even the central one obscured to the 2d person, the 4d person looking at a 3d cube can see all 27 blocks, even the central one obscured to us. So for a 4d person, a "surface" is in 3d. A person in 5d would see all of the 81 blocks making up the 4d hyper-rubic cube.

How would a hypercube look to us in 3d? How many blocks would be obscured from us if it was passed by us in all orientations? If you think about us in 3d passing a rubics cube 'square on' through a plane occupied by our 2d colleage, then he or she would see a square (albeit with the colours mysteriously changing up to three times). Do this in all three orientations, and our colleague will have seen all the blocks except the central block of our rubic cube.

What happens if our 4d colleague does the same for us -- pass a hyper-rubic cube through our 3d space, square-on, in all four orientations. We would see a rubic cube, albeit with its blocks mysteriously changing colours on the way through. But after all 4 orientations have been used, how many of the 81 blocks would have been obscured from our view. Is it only the one central block also obscured from our 4d colleague, or is it also the additional 8 central blocks of the cubes that make up the hypercube's surface?

My guess is that only the 1 central block will remain obscured from us 3d souls, and that we will see, but not all at once, what our 4d colleague sees in a glance. But I can't see how?

Bit of a rave. Hope it is clearer than mud :)

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#35

### Re: Rubik's Hypercube: Newsletter Challenge (August 2013)

08/13/2013 7:17 PM

This looks like a trick question to me. The obvious trap is to see this as a mathematical problem. It isn't. It's a reality check. No Rubik's cube I have ever seen was made up of cubes. The sum total number of cubes in a "typical" Rubik's cube is one. And you can see it, it's not hidden. Unless you have lost your cube along with your marbles under the bed somewhere. The component parts of a Rubik's cube can not be cubes. For Rubik's mathematical point to be made, there must be rules about how the component parts interact with each other. These rules are not consistent with making the components cubes and there can not be a hidden cube. A Rubik's cube with 5 SQUARES (they are not cubes) per edge would also have zero hidden cubes and only one cube in total. A Rubik's hypercube can not exist and the assumption that the said hypercube and the observer's perspective are 4D is meaningless. Of course, all else being equal, the hypercube will also have zero hidden hypercubes.

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#36

### Re: Rubik's Hypercube: Newsletter Challenge (August 2013)

08/14/2013 11:05 AM

Mr. English,

Look like cubes to me!
OK so maybe there isn't actually a center "cube", and the outer "cubes" have some additional and subtractive geometries, but I think you're getting stuck in the mud of reality. Free yourself for a moment, and focus on the boundaries of the question. I mean reality is great and all, but sometimes a little abstract thought is fun too.

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#37

### Re: Rubik's Hypercube: Newsletter Challenge (August 2013)

08/15/2013 12:51 AM

Don't stop me - Queen

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#39

### Re: Rubik's Hypercube: Newsletter Challenge (August 2013)

08/21/2013 1:19 AM

5 cubes per edge would have 27 hidden cubes.

3 hypercubes per edge would have 1 hidden hypercube.

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