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Polygon and platonic solids node problem

JT

7/18/2014 1:01:00 AM


1 2 1 2
3 4 3 4
1 2 1 2
3 4 3 4

Connecting squares there seem to be no problem having unique node names using just the individual 4 corners but what if it were interconnected cubes rather then squares?

That would not be possible with just four, i have a bit hard visualise it?

And what about a tetra, it also have four corners but only 3 nodes, is that possible?
2 Answers

JT

7/18/2014 3:17:00 AM

0

Den fredagen den 18:e juli 2014 kl. 03:01:10 UTC+2 skrev jonas.t...@gmail.com:
> 1 2 1 2
>
> 3 4 3 4
>
> 1 2 1 2
>
> 3 4 3 4
>
>
>
> Connecting squares there seem to be no problem having unique node names using just the individual 4 corners but what if it were interconnected cubes rather then squares?
>
>
>
> That would not be possible with just four, i have a bit hard visualise it?
>
>
>
> And what about a tetra, it also have four corners but only 3 nodes, is that possible?

Thinking about cubes pushing them together.
I want to avoid name collisions between corner when i connect them how can i calculate how many needed?

Any corner have 6 connection points plus itsalf, so with 7 different node names you can create a network of cubes without dissparance of names as you push them together?

I am a bit tired so i am not sure i am thinking straight here.

And how do you go on creating such a network of connected cubes giving names to the cube sides so they do not collide, trial and error?

Here is a really crazy idea, if you build such a network from inward to outward, and have your origo node starting from a cube, so each corner hold a node name 1-7, but do also hold a naturl number which you count up by corners as you push cubes together, so each corner now hold both a node name and also a natural number.

Given that your starting point is in middle of cube any number on the surface of the cube can be reached in (cube root/2), and you can encode the number by saving the nodes to it.
You create a lookup table for the naturals as you go along create the node tree.
But then again each node hold a value that must use 3-bits?

So is there any compression gained?

So any computer must hold the node tree lookup table to be able to decompress the number.
Any corner have 6 connection points plus itsalf, so with 7 different node names you can create a network of cubes without dissparance of names as you push them together?

I am a bit tired so i am not sure i am thinking straight here.

JT

7/18/2014 9:16:00 AM

0

Den fredagen den 18:e juli 2014 kl. 05:17:14 UTC+2 skrev jonas.t...@gmail.com:
> Den fredagen den 18:e juli 2014 kl. 03:01:10 UTC+2 skrev jonas.t...@gmail..com:
>
> > 1 2 1 2
>
> >
>
> > 3 4 3 4
>
> >
>
> > 1 2 1 2
>
> >
>
> > 3 4 3 4
>
> >
>
> >
>
> >
>
> > Connecting squares there seem to be no problem having unique node names using just the individual 4 corners but what if it were interconnected cubes rather then squares?
>
> >
>
> >
>
> >
>
> > That would not be possible with just four, i have a bit hard visualise it?
>
> >
>
> >
>
> >
>
> > And what about a tetra, it also have four corners but only 3 nodes, is that possible?
>
>
>
> Thinking about cubes pushing them together.
>
> I want to avoid name collisions between corner when i connect them how can i calculate how many needed?
>
>
>
> Any corner have 6 connection points plus itsalf, so with 7 different node names you can create a network of cubes without dissparance of names as you push them together?
>
>
>
> I am a bit tired so i am not sure i am thinking straight here.
>
>
>
> And how do you go on creating such a network of connected cubes giving names to the cube sides so they do not collide, trial and error?
>
>
>
> Here is a really crazy idea, if you build such a network from inward to outward, and have your origo node starting from a cube, so each corner hold a node name 1-7, but do also hold a naturl number which you count up by corners as you push cubes together, so each corner now hold both a node name and also a natural number.
>
>
>
> Given that your starting point is in middle of cube any number on the surface of the cube can be reached in (cube root/2), and you can encode the number by saving the nodes to it.
>
> You create a lookup table for the naturals as you go along create the node tree.
>
> But then again each node hold a value that must use 3-bits?
>
>
>
> So is there any compression gained?
>
>
>
> So any computer must hold the node tree lookup table to be able to decompress the number.
>
> Any corner have 6 connection points plus itsalf, so with 7 different node names you can create a network of cubes without dissparance of names as you push them together?
>
>
>
> I am a bit tired so i am not sure i am thinking straight here.

Sorry a square need at least 5 corner names to keep a one way path with no revisiting of square collision free?

I started to think about it in middle of night so it may all be wrong?
How many corner names needed to keep an outgoing oneway path from a start square collision free?

Observe there is no problem in that the start square pointing to a corner using the same name as itself, because it is oneway. And the squares are individual and their corners can not be revisited.

And how many would it take for a cube?