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Re: Number of needed nodes to create collison free network.

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

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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.
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> But I think i need some help to formulate the problem in a more coherent manner.
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> 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.
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> So i may need some help formulate the problem in a more coherent manner.
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> I will start using the easiest case a square.
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> A square is a unique individual that use different name for each corner.
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> The arrangement of the corners is free, a square and its corner can never be revisited.
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> 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.
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> 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.
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> How many corner names needed to create a collision free network.
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> 1. Triangle
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> 2. Tetrahedron
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> 3. Square
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> 4. Cube
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> And other polygons and platonic solids, is this group theory, computational complexity?
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> 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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Thread

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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