Naming and notations

Discussion of shapes with curves and holes in various dimensions.

Postby Keiji » Sat Jun 03, 2006 5:55 pm

HEE would be two solid squares.
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Postby moonlord » Sat Jun 03, 2006 6:39 pm

... that are the opposite faces of a cube.
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Postby Keiji » Sat Jun 03, 2006 8:41 pm

Correct.
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Postby bo198214 » Sat Jun 03, 2006 10:05 pm

Really sense make only the combinations
E...EH...H
But this you can then also express as the n-dim hypercube consisting of k-dim facets (if you have k 'E's and n-k 'H's). Or: the n-dim k-frame hypercube.

I mean with this notation you anyway can not represent every dimensional combinations of facets.
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Postby Keiji » Sat Jun 03, 2006 10:10 pm

Really sense make not your post ;)

(no offense, I just didn't understand)
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Postby bo198214 » Sat Jun 03, 2006 10:18 pm

if you have 'H's before 'E's then you get some strange objects like two solid squares or so. If the system should be complete I would assume we want to get all strange objects for example 4 vertices together with a solid square or so. But this can not be accomplished by the E/H system. So it is incomplete. And the main purpose was anyway I think to describe things like the x-frame y-dim something.
Even if you introduce round objects its usually only necessary to give dimensional thickness of the boundaries ...
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Postby Keiji » Sun Jun 04, 2006 12:58 am

bo198214 wrote:if you have 'H's before 'E's then you get some strange objects like two solid squares or so. If the system should be complete I would assume we want to get all strange objects for example 4 vertices together with a solid square or so. But this can not be accomplished by the E/H system. So it is incomplete.


The E/H system is only to allow for n-frame objects. The existence of "strange" objects is purely a side-effect. More importantly, these strange objects are all prismatic. If you can think of a prismatic "strange" object that can't be made using the E/H system, I'll give you an e-cookie. :P

Even if you introduce round objects its usually only necessary to give dimensional thickness of the boundaries ...


what?
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Postby PWrong » Mon Jun 05, 2006 3:25 pm

I think I understand this system now, at least when applied to 3D objects. It does seem pretty natural, although I think it should be used alongside RNS, rather than replacing it. Is there a list somewhere of all the objects up to 4D along with their CSG notation?

I do have a problem with the M, R and S operations. These could be made more general and less redundant by using matrices. Adding a vector translates the object, and multiplying by a matrix rotates/stretches the object.

I think I can work out the symbols for some of the toratopes. However, it might be a good idea to include the torus product as part of the notation.

Torus: ELM1[a]L = EL#EL
Start with a circle, translate it along the x-axis, then rotate.

Toracubinder: ELLM1[a]L = EL#ELL
Start with a sphere, translate it, then rotate

Toraspherinder: ELM1[a]LL = ELL#EL
Start with a circle, translate it, then rotate twice

Tritorus: ELM1[a]LM?[b]L = EL#EL#EL
Translate a torus along some axis, and rotate

Torinder: ELM1[a]LE == (EL#EL)E
Extrude a torus.

I'm not sure if (EL#EL)E is the same as EL#(ELE) or EL#ELE, hence the brackets. What is ELE anyway?
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Postby moonlord » Mon Jun 05, 2006 3:50 pm

ELE is a full cylinder. E gets a line, L transforms it into a disk and extrude it again to get a full cylinder. I'm not yet confident with the torus product so I can not confirm the rest :( ...
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Postby Keiji » Mon Jun 05, 2006 3:53 pm

PWrong wrote:I think I understand this system now, at least when applied to 3D objects. It does seem pretty natural, although I think it should be used alongside RNS, rather than replacing it. Is there a list somewhere of all the objects up to 4D along with their CSG notation?


Look on the individual polytope pages. I won't outright replace RNS with CSG, because RNS exists purely for rotopes - and besides, the CSG expression for a duocylinder is as yet unknown. Talking about a duocylinder, does it really have a hole? On this page it is projected as having a hole and it is the only 4D rotope that I can't visualize (other than the tiger, which is based on the duocylinder anyway :P )... And if it does, is it a pocket or a normal hole?

I do have a problem with the M, R and S operations. These could be made more general and less redundant by using matrices. Adding a vector translates the object, and multiplying by a matrix rotates/stretches the object.


You are absolutely right, but I know nothing about matrices (other than the fact that they generalize transformations). So that's up to you ;)

Those torus expressions are correct as far as I can see. As for your final question:

I'm not sure if (EL#EL)E is the same as EL#(ELE) or EL#ELE, hence the brackets. What is ELE anyway?


ELE is a solid cylinder: it is the same as LEE and EEL, albeit in different orientations.

(EL#EL)E would be an extruded torus. EL#(ELE) and EL#ELE would be the same object since # takes precedence, but I don't know what this would be since I don't really know how to spherate by a cylinder...
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Postby moonlord » Mon Jun 05, 2006 3:59 pm

Unfortunately, LEE is not a cylinder. That's... well... an artefact... :). L gives two points, which, after extrusion, give two squares... opposite faces of a cube.
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Postby PWrong » Mon Jun 05, 2006 4:24 pm

Talking about a duocylinder, does it really have a hole? On this page it is projected as having a hole and it is the only 4D rotope that I can't visualize (other than the tiger, which is based on the duocylinder anyway )... And if it does, is it a pocket or a normal hole?

I'm not sure. It might depend which form of the duocylinder you mean. I think the margin is topologically equivalent to a hollow torus, so it has both a hole and a pocket.

(EL#EL)E would be an extruded torus. EL#(ELE) and EL#ELE would be the same object since # takes precedence, but I don't know what this would be since I don't really know how to spherate by a cylinder...

The definition of the torus product is ambiguous on this point. There are two ways to spherate a circle by a cylinder, depending on how you rotate it. One is the torinder, while the other is something like a duocylinder, except one of the circles is replaced by a pair of concentric circles. It has the following equations:
R^2 < x^2 + y^2 < (R+h)^2
0 < z^2 + w^2 < r^2

The torus product is only strictly defined for A#k-sphere, where A can be any object. It might be interesting to look at it acting on other shapes.
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Postby Nick » Mon Jun 05, 2006 7:26 pm

I understand this notation, except for one thing...
What does the # mean?
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Postby Keiji » Mon Jun 05, 2006 8:36 pm

moonlord wrote:Unfortunately, LEE is not a cylinder. That's... well... an artefact... :). L gives two points, which, after extrusion, give two squares... opposite faces of a cube.


Sorry, what am I talking about; you're totally correct.

I'm not sure. It might depend which form of the duocylinder you mean. I think the margin is topologically equivalent to a hollow torus, so it has both a hole and a pocket.


I was talking about the duocylinder that was shown on the page I linked to, if there are other forms of it then that's even more confusing. hehe.
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Postby PWrong » Tue Jun 06, 2006 4:24 am

I understand this notation, except for one thing...
What does the # mean?

That's the torus product. It's the same concept as spheration. For instance, circle#circle is a hollow torus. You start with a large circle, then replace every point in it with a smaller circle. The plane of the small circle depends on the centre point.

I was talking about the duocylinder that was shown on the page I linked to, if there are other forms of it then that's even more confusing. hehe.

Duocylinder comes in three forms: 2D margin, a pair of 3D cells, and a 4D solid.
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Postby Keiji » Tue Jun 06, 2006 6:08 am

Then I would be talking about the 4D solid. So.. does it have a hole or not? :P
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Postby PWrong » Tue Jun 06, 2006 6:50 am

Ok, I don't think the 4D duocylinder would have a hole. If it did, it would probably be in the centre, but (0,0,0,0) is clearly part of the duocylinder.

The 3D form might be more difficult, because it doesn't include (0,0,0,0)

It might be useful to look at some curves that start at (0,0,0,0) and end outside the duocylinder, without crossing the duocylinder. If we can find one, we know the 3D duocylinder doesn't have a pocket.

The 3D duocylinder is the union of these sets:
A : x^2 + y^2 < 1 & z^2 + w^2 = 1
B : x^2 + y^2 = 1 & z^2 + w^2 < 1

Consider the curve r(t) = (0,0,0,t)
This is the line up the w-axis. It starts at (0,0,0,0) which isn't part of A or B.
When t=1, we get r(t) = (0,0,0,1), which is a member of A.
For t>1, r(t) is outside the duocylinder.
From symmetry, it looks like this every curve starting at the centre will have to pass through the object in order to escape.

So I think it's safe to say that the 3D duocylinder has a pocket, and it might also have a hole.
Obviously that's not a proof, partly because I don't have a strict definition for a hole (in terms of calculus).
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Postby PWrong » Tue Jun 06, 2006 7:01 am

the CSG expression for a duocylinder is as yet unknown

Isn't the duocylinder something like a rotated cylinder? Rotate one way, and you get a spherinder, rotate the other way and you get a duocylinder.
EELL is the spherinder, so maybe the duocylinder is ELEL. Or spherinder could just as easily be ELLE. That might be more sensible
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Postby Keiji » Tue Jun 06, 2006 11:22 am

PWrong wrote:
the CSG expression for a duocylinder is as yet unknown

Isn't the duocylinder something like a rotated cylinder? Rotate one way, and you get a spherinder, rotate the other way and you get a duocylinder.
EELL is the spherinder, so maybe the duocylinder is ELEL. Or spherinder could just as easily be ELLE. That might be more sensible


... I'm on a roll of stupidity right now... You're correct.

EELL and ELLE are spherinders in different orientations.
ELEL is a duocylinder.
LELE and LLEE are the curved face of a cylinder extruded into the fourth dimension, in different orientations.
LEEL is a pair of cylinders.
Last edited by Keiji on Tue Jun 06, 2006 6:33 pm, edited 1 time in total.
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Postby moonlord » Tue Jun 06, 2006 1:03 pm

LLEE seems to me to be a cylindrical surface, extruded in W. That is by no means a pair of cylinders :?

BTW, ELEL seems to be the tetraframe duocylinder. Am I right?
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Postby Keiji » Tue Jun 06, 2006 6:34 pm

I truly am on a roll of stupidity. Maybe it's my exams :roll: :mrgreen:

I fixed my post anyway. And ELEL is indeed the tetraframe duocylinder.
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Postby moonlord » Tue Jun 06, 2006 6:49 pm

Although the thread is about notations, I see we're attacking the duocylinder again. There you go:

Let's take it otherwise. I'm reffering to cartesian product in the following.

A tetraframe duocylinder is disk x disk.
A triframe duocylinder is disk x circle or circle x disk, I don't think it matters.
A diframe duocylinder is circle x circle.

I believe we can gather information about holes from these.

A tetraframe duocylinder has no hole of any kind. Disks are full, and we don't spherate anything.
A triframe duocylinder needs some more thinking. I'm not yet sure which product is the correct one.
A diframe duocylinder is even more problematic.

Now you've got some fresh ideas. Maybe you can help.
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Postby Keiji » Tue Jun 06, 2006 7:35 pm

Hmm, so what exactly is the cartesian product?
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Postby bo198214 » Tue Jun 06, 2006 8:23 pm

For sets A of R<sup>k</sup> and B of R<sup>m</sup> the flat cartesian product AxB is the set of all tupel (x<sub>1</sub>,...,x<sub>k</sub>,x<sub>k+1</sub>,...,x<sub>k+m</sub>) where (x<sub>1</sub>,...,x<sub>k</sub>) is element of A and (x<sub>k+1</sub>,...,x<sub>k+m</sub>) is element of B.
So the flat cartesian product is from P(R<sup>k</sup>) x P(R<sup>m</sup>) to P(R<sup>k+m</sup>). Where P(X) is set of all subsets of X.

For example take two lines on R<sup>1</sup> as the intervals [0,1], then
[0,1]x[0,1] is the set of all (x,y) such that 0<=x,y<= 1 and that is the full square. Or take a circle and a line and you get a cylinder. Your responsibility for higher dimensions ;)
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Postby Keiji » Wed Jun 07, 2006 12:10 am

...

You lost me at R<sup>k</sup>. :shock:
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Postby Marek14 » Wed Jun 07, 2006 6:42 am

moonlord wrote:Let's take it otherwise. I'm reffering to cartesian product in the following.

A tetraframe duocylinder is disk x disk.
A triframe duocylinder is disk x circle or circle x disk, I don't think it matters.
A diframe duocylinder is circle x circle.

I believe we can gather information about holes from these.

A tetraframe duocylinder has no hole of any kind. Disks are full, and we don't spherate anything.
A triframe duocylinder needs some more thinking. I'm not yet sure which product is the correct one.
A diframe duocylinder is even more problematic.

Now you've got some fresh ideas. Maybe you can help.


Both product are correct for triframe duocylinder. Each duocylinder has two of these. If the circles that form it have different radii, the triframe duocylinders are different.

Do they contain a hole? Maybe we can answer this question by looking at the slices.

Duocylinder when sliced looks like a circle which stretches into a cylinder, then shrinks back. The triframe duocylinder would have the circle empty and only have a mantle of the cylinder filled (in one orientation). We can put a plane through the middle of duocylinder which will show in our slicing as an unmoving line through the centers of cylinder's circles. So this means there's a plane which doesn't pass through one specific tricylinder, but cannot be "pulled out" without intersecting it, which means that there is a hole.
It can be seen even better in the other slice of triframe duocylinder, which starts as a full disk, and becomes two disks which separate in thrid dimension, then merge back together.

By the way - I already asked this, but where does thie EL notation come from? I haven't seen it anywhere...
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Postby PWrong » Wed Jun 07, 2006 8:21 am

You lost me at R<sup>k</sup>.

R<sup>k</sup> is ordinary k-dimensional (euclidean) space.

By the way - I already asked this, but where does thie EL notation come from? I haven't seen it anywhere...

I think Rob invented it. I don't know where he first started using it.
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Postby bo198214 » Wed Jun 07, 2006 8:24 am

Rob wrote:You lost me at R<sup>k</sup>. :shock:


R<sup>k</sup> is simly the set of k-tupels, i.e. all (x<sub>1</sub>,...,x<sub>k</sub>) where x<sub>1</sub>,...,x<sub>k</sub> are real numbers.

I wonder whether the the triframe duocylinder is disk x circle united with circle x disk ...
For example the normal 3d cylinder is (circle x interior of line) u (disk x end points of line).
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Postby PWrong » Wed Jun 07, 2006 8:32 am

A triframe duocylinder is disk x circle or circle x disk, I don't think it matters.

The cartesian product is supposed to be commutative. That is, disk x circle = circle x disk
However, you might have the circles and disks in different planes.

One cell of triframe duocylinder is disk(x,y) x circle(z,w), while the other cell is disk(z,w) x circle(x,y)
The total triframe duocylinder is the union of the two cells.
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Postby bo198214 » Wed Jun 07, 2006 8:41 am

PWrong wrote:The cartesian product is supposed to be commutative. That is, disk x circle = circle x disk

:shock: in my definition its not commutative and never heard of a commutative cross product (not even the completely different vector crossproduct is commutative). It looks also then that you contradict in your post.

Admittetly the result of both A x B and B x A maybe geometrical the same object (btw A and B are by definition in a different subspace). At least its then not the triframe duocylinder, which is the union of two such geometrical identical objects.
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