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Groups > comp.lang.javascript > #25431
| Newsgroups | comp.lang.javascript |
|---|---|
| Date | 2014-07-18 04:45 -0700 |
| References | <717c28f1-3871-4309-99c8-c4b632f61126@googlegroups.com> |
| Message-ID | <373e0adc-7815-4b3f-9a46-d71fc5073bfd@googlegroups.com> (permalink) |
| Subject | Re: Number of needed nodes to create collison free network. |
| From | jonas.thornvall@gmail.com |
Den fredagen den 18:e juli 2014 kl. 12:15:35 UTC+2 skrev jonas.t...@gmail.com: > I was a bit tired tonight thinking about this, and very diffuse in my problem statement. I try be a bit more coherent below. > > > > But I think i need some help to formulate the problem in a more coherent manner. > > It is about how many corner/node names needed to create a collsion free network. Warning i am not that good formulate the actual problem. > > > > So i may need some help formulate the problem in a more coherent manner. > > > > I will start using the easiest case a square. > > > > A square is a unique individual that use different name for each corner. > > The arrangement of the corners is free, a square and its corner can never be revisited. > > > > Now i want to build a oneway network out from a cental starting square. I push squares together, creating outward nodes from a central square that will be collision free. ***You only push together corners holding same name*** thus a interconnected corner/node is named 1,2,3,4,5... and so on. > > > > To be collision free means that a corner can not point to two corners holding same name, but it can itself hold the same name as a corner it pointing to because it is a oneway path network. > > > > How many corner names needed to create a collision free network. > > > > 1. Triangle > > 2. Tetrahedron > > 3. Square > > 4. Cube > > And other polygons and platonic solids, is this group theory, computational complexity? > > > > Maybe it is self evident but i just do not see it. Kind of an update, i just realised that every node of a cube could hold 6 naturals depending on the taken path. (Well given the above and my naive assumption that we can create a cubed network with no collision just using 6-7 names so storing a node 3 bits? Is it true that an arrangement arround a centralised cube where every named corner intersection/nodes can hold 6 different natural number depending on the path we take and as we add cubes we add up natural numbers. Now is it true that from this *centralised* cube any natural number arranged along the cube path can be reached in 3th root/(2*6)+1 **steps** using 3 bits? Well the 2 is because the centralised arrangement and the 6 the number of natural numbers stored at each node.
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Number of needed nodes to create collison free network. jonas.thornvall@gmail.com - 2014-07-18 03:15 -0700
Re: Number of needed nodes to create collison free network. jonas.thornvall@gmail.com - 2014-07-18 04:45 -0700
Re: Number of needed nodes to create collison free network. Ben Bacarisse <ben.usenet@bsb.me.uk> - 2014-07-18 12:57 +0100
Re: Number of needed nodes to create collison free network. jonas.thornvall@gmail.com - 2014-07-18 05:30 -0700
Re: Number of needed nodes to create collison free network. Ben Bacarisse <ben.usenet@bsb.me.uk> - 2014-07-18 13:43 +0100
Re: Number of needed nodes to create collison free network. jonas.thornvall@gmail.com - 2014-07-18 08:49 -0700
Re: Number of needed nodes to create collison free network. jonas.thornvall@gmail.com - 2014-07-18 11:10 -0700
Re: Number of needed nodes to create collison free network. Ben Bacarisse <ben.usenet@bsb.me.uk> - 2014-07-18 20:00 +0100
Re: Number of needed nodes to create collison free network. jonas.thornvall@gmail.com - 2014-07-18 12:28 -0700
Re: Number of needed nodes to create collison free network. Ben Bacarisse <ben.usenet@bsb.me.uk> - 2014-07-18 21:32 +0100
Re: Number of needed nodes to create collison free network. "Michael Haufe (TNO)" <tno@thenewobjective.com> - 2014-07-19 04:41 -0700
Re: Number of needed nodes to create collison free network. jonas.thornvall@gmail.com - 2014-07-19 07:37 -0700
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