RNS and product notations

[PWrong]

Did we just arbitrarily define 1 = line, 2 = circle, etc?

Pretty much. The digit notation and group notation together are called RNS notation. Putting two objects together gives the cartesian product of the two objects. I'm still not really sure what putting brackets around an object does. There's a way to convert directly from RNS to parametric equations or to implicit cartesian equations, but not to "product notation".

moonlord: Split from "Definition of the taper operation"

[bo198214]

I'm still not really sure what putting brackets around an object does.

And thats what makes me really wonder. There are extended threads using this notation and it seemes that nobody really knows what it means!
(Rob thought it was bending the object and gluing cells together, no its spheration, for which there is no clear definition either. I mean how can you guys count rotopes, determine volume, derive parametric equations if everybody has its own definition of the parenthesis?!)

So I want to have a clear definition of these parentheses, on which everybody agrees here. And then this should be put into the wiki. The wiki consists anyway of too much discussion of too undefined things.

[PWrong]

no its spheration, for which there is no clear definition either

Yes there is. It's at the end of this thread
For some reason, noone replied on that thread except wendy.

I mean how can you guys count rotopes, determine volume, derive parametric equations if everybody has its own definition of the parenthesis?!

The parenthesis can be defined using the conversion to cartesian equations. The only thing we don't know is how to convert from RNS to product notation.

[bo198214]

no its spheration, for which there is no clear definition either

Yes there is. It's at the end of this thread

Sorry, but I dont see it there. And I am also a bit unwilling to read a whole thread with "your notation" and "my notation" only to have again uncertainty about the base terms.

Would you be so nice and put a final definition of spheration into this thread together with a (maybe parametric equation) definition of the RNS notation? Or even better into the wiki with a link here.

[PWrong]

Essentially, you take the normal space of A at each point, combined with a few extra dimensions. These are the new axes. Now put a B at every point in A, aligned along these axes.

Example: circle#sphere
A circle is a 1D shape in 2D space, so it has a normal vector at each point(other objects may have a normal plane or some other space). So we have a 1D normal space. A sphere requires 3 dimensions, so we need two more.

So at each point in the circle, we place a sphere aligned along three axes: the normal vector of the circle, the z-axis and the w-axis.

Here's a strict definition. Let A be a kD object in nD space, let B be an mD sphere, and let N be the normal space of A. Then,
A#B = {u + v| u E A(e₁ ... e_k), v E B(N, e_k+1 ... e_k+m-n) }

The e_i are basis vectors, and E is the "element" symbol.

[bo198214]

Hm, a strange definition of spheration where the word "spheration" does not occur. So what has this # now todo with the parentheses in RNS notation (as when Rob said the parentheses mean spheration)?

[Keiji]

(x1) = x # circle.

But, how is the tiger, (22), circle x circle # circle? I don't get it.

[bo198214]

Alas, and I thought I am the only one *nudge*

Hm, but then would (11) be the cylinder? No Rob, this can not be.

So do I see it right that the RNS notation is basicly a notation to define parametric functions where the corresponding shape is yielded by its zero set?

So that (F₁...F_n) is defined as F₁²+....+F_n² - R², where R is a constant?

The equation for the tiger is:
(sqrt(x^2+y^2) - r1)^2 + (sqrt(z^2+w^2) - r2)^2 < r3 ^2

Rob, cant you make a slicing sequence for the tiger (in a new thread)?

[PWrong]

Hm, a strange definition of spheration where the word "spheration" does not occur.

Sorry, forgot to mention that spheration is the same as the torus product. A#B is A spherated by B. Note that B is always a sphere of some dimension.

So that (F1...Fn) is defined as F12+....+Fn2 - R2, where R is a constant?

Close. It's sqrt(F1^2+....+Fn^2) - R

[Marek14]

(x1) = x # circle.

But, how is the tiger, (22), circle x circle # circle? I don't get it.

Well, that can be explained by normals too: circle x circle is a 2D shape in four dimensions, so at every point, it has normal 2D plane (based on old mathematical equality 2+2=4 ). I did not check this, but I think that this plane is determined by normal vectors of both circles at the point. And it's this plane where you put the circle.

Otherwise, I agree that tiger looks a bit weird when written like spheration. Personally, I consider spheration a proper subset of parenthesis notation, where the outermost set of parentheses contains at most one non-1 figure.

[Keiji]

The equation for the tiger is:
(sqrt(x^2+y^2) - r1)^2 + (sqrt(z^2+w^2) - r2)^2 < r3 ^2

Rob, cant you make a slicing sequence for the tiger (in a new thread)?

I'll post them in the "slicing toratopes with hyperplanes" thread.

[PWrong]

I can't remember if I've noticed this before, but I've found a nice way to group toratopes together. I group them by the number of levels of nested brackets. I've also put the product notation for each shape. Maybe this will help us find a rule for converting RNS to product notation.

I occasionally use brackets in the product notation, because # and x aren't associative. (2x2)#2 is the tiger, while 2x(2#2) is circle x torus (which is a 5D shape)


4D: 1 + 4 + 4 + 1 = 10
1111 = 1x1x1x1

211 = 2x1x1
22 = 2x2
31 = 3x1
4 = 4

(21)1 = (2#2)x1
(211) = 3#2
(22) = (2x2)#2
(31) = 2#3

((21)1) = 2#2#2

5D: 1 + 6 + 10 + 6 + 1 = 24

11111 = 1x1x1x1x1

2111 =2x1x1x1
221 = 2x2x1
311 = 3x1x1
32 = 3x2
41 = 4x1
5 = 5

(21)11 = (2#2) x1x1
(211)1 = (3#2) x1
(2111) = 4#2
(22)1 = ((2x2)#2) x1
(21)2 = (2#2)x2
(221) = ?
(31)1 = 2#3 x1
(311) = 3#3
(32) = (3x2)#2 ?
(41) = 2#4

((21)1)1 = 2#2#2 x1
((21)11) = (3#2)#2
((211)1) = ?
((22)1) = ((2x2)#2)#2 ?
((21)2) = ?
((31)1) = ((2#2)#3) ?

(((21)1)1) = 2#2#2#2

Can anyone fill in or correct the question marks?

[moonlord]

One should note that X (X>1) is the same with (111..1) - X times.

So 11 is the square, (11) = 2 is the disk; 111 is the cube, (111) is the sphere, (11)1 is the cylinder, ((11)1) is the torus. Although I fail to see the transformation...

[Keiji]

(221) = 2#(2x2)

((211)1) = 2#(3#2)

^ using my "theory" that (x1) = 2#x, given that x != 1

[PWrong]

Then I would introduce the brackets [] for rectangular compositions and the parentheses () for spherical composition.

We used to use square brackets for RNS. They turned out to be redundant.

Otherwise the notation is ambigous.

We can directly convert RNS to cartesian equations, so it's definately not ambiguous. The only problem we have is converting RNS to product notation. The last thing we need is an entirely new notation.

(221) = 2#(2x2)
((211)1) = 2#(3#2)

That's no good. I've said before that A#B only exists if B is a sphere.

EDIT: I just realised I made the same mistake a couple of times above. I've fixed them now.

[Keiji]

(221) = 2#(2x2)
((211)1) = 2#(3#2)

That's no good. I've said before that A#B only exists if B is a sphere.

EDIT: I just realised I made the same mistake a couple of times above. I've fixed them now.

Then, (2x2)#2 and (3#2)#2 instead...

[Marek14]

(221) = (2x2)#3
(32) = (3x2)#2 ? Yes
((211)1) = (2#3)#2
((22)1) = ((2x2)#2)#2 ? Yes
((21)2) = (2#2)#3
((31)1) = ((2#2)#3) ? Nope, (3#2)#2

Although something seems fishy here... what direction do you define spheration in? You have (31)1 as 2#3 x1, while it should be 3#2 x1, AFAIK

----

From Bo's post after the topic was split (I wish phpbb could split posts...) ~Rob

That's no good. I've said before that A#B only exists if B is a sphere.

Come on Paul! You defined # for arbitrary shapes. If I take the extremly simple example of 1#A in 3d, where A is some arbitrary figure in 2d, whats the problem of doing it?!

[bo198214]

Ok, I agree with Robs assertion that
(A1) is equal to A # circle, for A being an RNS parametric function.

This is because for RNS parametric functions F it seems that F=eps, is all points of F=0 translated a bit along the normal space of F=0 (for eps some small constant). We have a natural 1 dimensional normal space in each point for those F. Even for non-smooth shapes like [11] that have in certain points no normal space we get by (F1) a natural definition of F # circle. Similarly we have then a natural definition for F # square by [F1].

This works also for arbitrary spheration, ie.
(A1...1) with d-times 1 is equal to A # S_d. Even for d=0!

Any comments on this, or was this already known long before, or are there counterexamples?

And now as I think of it, the connection between parenthesis notation and # reveals as quite clear:
(A₁....A_n) is always equal to (A₁x....x A_n) # S_n-1

For an example take the tiger, or even we can use a 2 dimensional example. Consider ((1)(1)) - where (1) is by (parametric) definition two points, i.e. the 0 dimensional sphere. The parametric equation is
sqrt((|x|-r1)^2 + (|x|-r2)^2) - r3 = 0
these are 4 circles with radius r3 at (r1,r2), (-r1,r2), (r1,-r2) and (-r1,-r2). But its product notation (S₀ x S₀) # S₁ is exactly the same S₀ x S₀ are 4 points and each point is then replaced by a circle (in its normal space).

So I would guess the most general form
(A₁....A_n1...1) with d 1's is always equal to (A₁x...xA_n) # S_n+d-1

[Marek14]

Ok, I agree with Robs assertion that
(A1) is equal to A # circle, for A being an RNS parametric function.

This is because for RNS parametric functions F it seems that F=eps, is all points of F=0 translated a bit along the normal space of F=0 (for eps some small constant). We have a natural 1 dimensional normal space in each point for those F. Even for non-smooth shapes like [11] that have in certain points no normal space we get by (F1) a natural definition of F # circle. Similarly we have then a natural definition for F # square by [F1].

This works also for arbitrary spheration, ie.
(A1...1) with d-times 1 is equal to A # S_d. Even for d=0!

Any comments on this, or was this already known long before, or are there counterexamples?

Well, I was using it like this from the beginning.

[PWrong]

(221) = 2#(2x2)
((211)1) = 2#(3#2)

That's no good. I've said before that A#B only exists if B is a sphere.

EDIT: I just realised I made the same mistake a couple of times above. I've fixed them now.

Then, (2x2)#2 and (3#2)#2 instead...

(2x2)#2 is the tiger.

Come on Paul! You defined # for arbitrary shapes. If I take the extremly simple example of 1#A in 3d, where A is some arbitrary figure in 2d, whats the problem of doing it?!

This is just a pair of A's. The orientation of each A is arbitrary. If A is a square, then 1#A could look like a pair of squares or a pair of diamonds, or a square and a diamond.

Maybe we should be trying to convert backwards, from product to RNS. That might be easier since the torus products are easier to visualise.

[Keiji]

(221) = 2#(2x2)
((211)1) = 2#(3#2)

That's no good. I've said before that A#B only exists if B is a sphere.

EDIT: I just realised I made the same mistake a couple of times above. I've fixed them now.

Then, (2x2)#2 and (3#2)#2 instead...

(2x2)#2 is the tiger.

That can't be right.

2x2 = duocylinder, 4D shape

4D shape # 2D shape = 5D shape

...?

The orientation of each A is arbitrary. If A is a square, then 1#A could look like a pair of squares or a pair of diamonds, or a square and a diamond.

In RNS, yes. However, CSG notation defines orientation.

So: EC₀#EEC₁ would definately be a pair of monoframe squares. And EC₀#EDC₁ would definately be a pair of monoframe squares rotated 45 degrees in xy.

[PWrong]

That can't be right.

2x2 = duocylinder, 4D shape

4D shape # 2D shape = 5D shape

No, duocylinder is a 2D shape that exists in 4D space (just like a helix is a 1D shape in 3D space). It's normal space is (4-2) = 2D, which is enough room to spherate by a circle without increasing the dimension.

[Keiji]

Marek14 posted here how to properly convert RNS to product notation.

[PWrong]

I noticed that, I just hadn't replied yet. I'll quote the whole thing here for reference.

1. Parentheses evaluate from inside out.
2. Parentheses of form a1111... with b 1's evaluate to a#(b+1)
3. Other parentheses containing any 1's evaluate to a#(b+k) where a#b is evaluation of the same parentheses without 1's, and k is the number of 1's.
4. Parentheses containing terms a,b,c etc. with at least two terms and none of them equal to 1 evaluate to (a x b x c x ...)#n where n is the number of terms.

Now I'll look at them one by one.

2. Parentheses of form a1111... with b 1's evaluate to a#(b+1)

I agree with this one.
Example: (211) = 2#4, (3111) = 3#5

3. Other parentheses containing any 1's evaluate to a#(b+k) where a#b is evaluation of the same parentheses without 1's, and k is the number of 1's.

I think I understand this one. So (32) = (3x2)#2 implies (321111) = (3x2)#(2+4) = (3x2)#6, correct?

4. Parentheses containing terms a,b,c etc. with at least two terms and none of them equal to 1 evaluate to (a x b x c x ...)#n where n is the number of terms.

So this would mean that (4332) = (4x3x3x2)#4.

I like all these rules. The dimensions always match up, and they all seem to work for all the low dimensional shapes. I think you've solved the problem Marek .

[PWrong]

I just realised Marek's rules don't solve the problem entirely. They don't help us convert ((21)2), for instance. Marek said it would be (2#2)#3. I disagree with this. If we substitute T = (21), then rule 4 gives us:
(T2) = (Tx2)#2
and since T = 2#2, we get
((21)2) = ((2#2)x2)#2

[Marek14]

I just realised Marek's rules don't solve the problem entirely. They don't help us convert ((21)2), for instance. Marek said it would be (2#2)#3. I disagree with this. If we substitute T = (21), then rule 4 gives us:
(T2) = (Tx2)#2
and since T = 2#2, we get
((21)2) = ((2#2)x2)#2

Ah - that was my mistake. You are correct.

EDIT: Actually, you use my rules to get the proper form - so the rules were right, I just made a mistake in their application.

[bo198214]

I think you've solved the problem Marek .

I hate doing that, but for the records:
I have found the result same time with Marek.
And I am really pissed off by that kind of group dynamics.

[PWrong]

EDIT: Actually, you use my rules to get the proper form - so the rules were right, I just made a mistake in their application.

True, but I had to assume that the rules apply to objects with more than two nested brackets.

I hate doing that, but for the records:
I have found the result same time with Marek.

Sorry Bo. I saw you said that, but I didn't understand the rest of your post. Congrats to you too.

And I am really pissed off by that kind of group dynamics.

What? :?

[moonlord]

And I am really pissed off by that kind of group dynamics.

What? :?

You know, the sheep effect...

[PWrong]

I'm being a sheep because I agree with Marek? I just think the rules look valid. Hopefully someone can prove them and we'll know for sure.