Wings as a math visualization tool...

[Dimitri]

Wings may have an excellent use as a math visualization tool too. Below is one such math problem I did conceive and visualize for putting it in Facebook, so that there may begin a discussion about finding a formula to solve it.

The question is: what is the formula that can give us the way to find how many cubes may be extracted from cubes constructed from whole-smaller cubes. Keep in mind that there are many cubes, inside the total cubes, that interpenetrate each other (as it is illustrated partly in the small image series of the first attached image). To explain it in a more simple way: let say that we have a cube constructed from four smaller cubes (as in the second attached image below), there we have the four small cubes + the one that is constructed from all of them = 5 cubes. So, how many cubes would we have in case of a whole cube which is constructed from 512 whole –undivisible- smaller cubes (as the one in the upper right corner in the first image)?

It is a problem that I did conceive it but I still did not be able to find a solution. So, if anyone has an idea let share it. If not, let be content with the enjoyment of the visuals. : - )

(the rendering is made with Blender's internal renderer)

[micheus]

let say that we have a cube constructed from four smaller cubes (as in the second attached image below), there we have the four small cubes + the one that is constructed from all of them = 5 cubes

Dimitri, is that right?
I see 8 small cubes + 1 from all => 9

[Avros]

"To explain it in a more simple way: let say that we have a cube constructed from four smaller cubes (as in the second attached image below), there we have the four small cubes + the one that is constructed from all of them = 5 cubes. "

Would not the answer for the second attachment be 9 as it shows eight cubes all up and then the total as another cube?

Then you want a formula that will add consecutive cubes as it expands outward...

[micheus]

Dimitri, assuming my comment is right I will try a shot.

So, by considering a cube composed by 4 units (small cubes) and using Wings3d to check the progression of the total amount I got this:

and it means that:
- the big cube composed of 4 smaller cubes of dimensions can group 64 smaller cubes inside of it;
- if we assume the smaller cube has 2 cubes of dimension, then the big cube can group 27 cubes inside of it;
- if we assume the smaller cube has 3 cubes of dimension, then the big cube can group 8 cubes inside of it;
- if we assume the smaller cube has 4 cubes of dimension, then the big cube can group only 1 cubes inside of it;

The total amount of cube representations possible are 100 (64+27+8+1).

As we can notice, we were talking about 4³+3³+2³+1³

Assuming that n is equivalent to the amount the smaller cube that defines the initial cube we can use the equation (ref.):

If everything I wrote is right, now the answer for your question is easy to find:
Tc=((512(512+1))/2) => 17247043584!!!

[Dimitri]

Sorry, it was my fault regarding the second image... indeed, it does not make 5 cubes, it makes 9. I became confused when writing the text of the post. : - )

Micheus, to be honest I did not understand your presentation and, of course, I can not be sure as to its solution’s being correct or not. Besides, my math knowledge is very poor… whenever I approach such problems I do approach them mostly by an arithmetical way than in a mathematical way (I mean not by formulas, equations etc)… and using as an aid visualizations, as much as it is possible.

As I said, maybe your solution is ok, I can not be sure (I did not be able to find any solution to it yet using my own ways)… I do have, however, an impression that you did not calculate the overlapping cubes (the ones in the in between areas). So, for being more clear I am putting a 2D diagram below. The total quadrangles that can be derived from the initial-big quadrangle (‘a’), including the overlapping ones (those in images ‘d,e,f,g,h.i,j’), plus the initial-big quadrangle (‘a’) are 30 (a=1+b=16+c=4+d=2+e=2+f=1+g=1+h=1+i=1+j=1=30).

As you see there are not only the ‘regularly’ arrayed quadrangles (as those in images 'b' and 'c') in the table, there are also the overlapping ones, that’s to say the ones located between the regularly arrayed ones. So, if you try to find those too in a table constituted from much more quadrangles than the 16 of the diagram below, finding and calculating the overlapping ones becomes very difficult. Imagine, for example, a big quadrangle constituted from 64 small quadrangles… how many quadrangles can be found in such a table?

When the question is put forth in a two dimensional way -and having a quadrangle constituted from a total of a few smaller quadrangles- is easy to find the solution but… when it comes to the three dimensional plane and instead of a quadrangle you have a cube, things become too much complicated. : - )

[micheus]

Dimitri, let's try to explain my approach...

I started thinking to use Combination, but then I realize it can't be applied to this situation. Then, I start to observe the evolution of the amounts for each possible combination.

So, that was what you proposed:

"Mega cube" dimension = Md = 4 smaller cubes (as you proposed)

a) how many cubes of 1 smaller cube size can be place into the Mega Cube?
b) how many cubes of 2 smaller cube size can be place into the Mega Cube?
c) how many cubes of 3 smaller cube size can be place into the Mega Cube?
d) how many cubes of 4 smaller cube size can be place into the Mega Cube?

The a) and d) answers are obvious, of course, but to get the other values I used the same approach of you by making the combinations one-by-one by hand. So, I think I have used the overlapped options too. Take a look at the pictures:




based on that I found these answers:
a) 64; b) 27; c) 8; d) 1;
and thinking of them, I noticed that 64=4³; 27=3³; 8=2³; 1 that could be 1³ (why not ).

that was the way I did the relationship between these data:
Value = (Md - size tested + 1)³
a) size tested = 1 => (4 - 1 + 1)³ => 4³
b) size tested = 2 => (4 - 2 + 1)³ => 3³
c) size tested = 3 => (4 - 3 + 1)³ => 2³
d) size tested = 4 => (4 - 4 + 1)³ => 1³

The equation I put in the previous represent the way to determine the sum of these values for any Md. That exponent ³ is related to the dimension used and thinking about your last example, its as two dimensional representation. Now, we could say that the same role would be applied to your two dimensional squares: 4²+3²+2²+1² => 30
but, the equation for calculate the sum of squares for any dimension (n) will be this one:

for n=64 (your example), Tc=(64x(64+1)x(2x64+1))/6 => 89440 !!!

Even if I'm wrong, that was a very interesting challenge Dimitri. Thanks.

[Dimitri]

Yes, now it is ok Micheus... with your new visual demonstration it is clearly evident that you got the question. And, also, it seems you did find the way to the answer. I am not so sure yet however, I have to inspect the logic you followed... but it seems that it is working, indeed. Happy to see that. I will think on it. : - )

It is true that I do like geometry problems but the main reason I made this post was not for finding a solution to the presented problem... it was mainly for putting forth the abilities that 3D modelling offers for their presentation (as is so evident from your visuals too... they are very cleverly conceived, really).

Thank you for your so inspired effort...

(btw, not 'quadrangles' but 'squares': my so poor English... : - )