From: lunatic@netcom.com (Lunatic Johnathan Bruce E'Sex)
Subject: Re: C3 Network Questions......
Organization: Netcom; Renegade User
Date: Thu, 17 Dec 1992 00:50:15 GMT

In article <BzB9HA.IGB@undergrad.math.waterloo.edu> a2chao@undergrad.math.waterloo.edu (Allen Chao) writes:
>In article <1992Dec14.221327.1231@netcom.com> lunatic@netcom.com (Lunatic Johnathan Bruce E'Sex) writes:
>>          A
>>         /|\
>>       /  |  \
>>     /     \   \
>>   /        \    \
>>  B---ECM---C-----D
>>
>
>Actually, if there were communications like so, why would there be need of a
>C^3 Master?  For example, a link shouldn't exist between C and D or B and C,
>because I doesn't go thru the Master at A.  Paths should exist like this:
>
>          A
>         /|\
>        / | \
>       B  C  D
> 
>...and that's all.  That's why there's a 5 ton (or whatever) computer inside
>unit A.  Link C to D works C-A-D.  Link B to C works the same.  Now don't
>you think that that's reasonable? 8)  (Least I think so...)
 
    ___
     |he way I see it, all the C^3 slaves are constantly in contact
 with the master.  They querry the master as to what the closest node
 of the C^3 system is to their target.  The master then tells them,
 and sends a signal to the other C^3 slave to open up its communications
 path to send data directly to the first C^3 slave.  Using this method,
 the master computer is merely a coordinator, not a router.  It doesn't
 need to handle the high volume of tactical information itself, it just
 directs the slaves to send it between each other.  Destroying the
 master computer eliminates all coordination between the links.


>(But wait a minute, you said the same thing that I said in the next paragraph
>and contradict yourself...)

    ___
     |hat's the illustration of an alternate method (everything
 after the "==>").
 

>> Say, 2 tons and 2 critical spaces instead of 1 each.  That's an
>> equivalent sized network (5+1+1+1 = 2+2+2+2 = 8 total).  I think I'll
>> add that to Conbattler....  (:  ( ==> Standard C^3 can route around
>> ECM.  C^3 Direct [2 tons/crits] cannot, but doesn't require a master
>> command computer which can break the whole network if destroyed.)
>>
>	The 'by-the-book' C^3 should be the one that cannot reroute,
>since all info must pass thru the main unit. (One good tactic is to cover
>the C^3 main with ECM...)  Your alternative system should be the one to
>be able to reroute, since each unit can do the same job as the other units.
 
 
    |\|o, the alternate system couldn't reroute because each node only
 knows about the nodes it is directly connected to, not the ones that
 they are connected to, as well.  It doesn't have "the big picture"
 that a master computer would have.  You can reroute with a master
 because it can tell them "Ok, you at node B pump your data to node A,
 and you at node A pump your data to node C."  Since the master can
 send and receive data like any other node, it could be part of the
 chain or not, but it doesn't HAVE to be.

     _
    /-\nyway, if all the data MUST go through the master, as you
 describe, then you'd have a hell of a time actually playing the
 game and trying to tell the opposing player why his intervening
 ECM can break one C^3 link but not another, without giving away
 the location of the C^3 master.  And at most it would take him
 only two steps to figure it out himself.  Using the example:
 "Ok, intervening ECM doesn't break the B-C link, so neither of
 those has the master.  Now all I have to do is get between one
 of those mechs and either A or D to find out which one has it."

-- 
   _______________________________________________________
  /         -= Lunatic Johnathan Bruce E'Sex     (:      /
 /  lunatic@netcom.com  GEnie: LUNATIC  CI$: 76170,672  /
/______________________________________________________/