8bdd5bb0   
   From: D.Goncz-A.A.S.M.E.T.-CPS-NPI1659534493@replikon.net   
      
   Good morning Class...   
      
   I'm still at it!   
      
   I use the microphone now because my hands are shaky...   
      
   It works okay...   
      
   From a difficult family history comes an obsession with reproduction, and a   
   commitment to redirecting that natural urge to something of more altruistic   
   character.   
      
   I've studied the problem of the self-reproducing machine since 1979, after   
   discussing it with Dr Rota who said it was a good problem worthy of   
   studying but we couldn't find funding so I left MIT....   
      
      
   I found a way to get the funding!   
      
      
   I wanted to share the details with you all first...   
      
      
   If we presume that there could be:   
      
   A demand vector,   
      
   d   
      
   Comprising in what is called technological order,   
      
   The demands for each of the components comprising or making up a   
   self-reproducing factory in which the workload of reproduction was shared   
   among multiple machines according to the following idea:   
      
   Since any machine can usually make at least one of its own parts but never   
   all of them we can guess that there would be a fraction,   
      
   1/n   
      
   Approximating the degree to which on the average each machine can make not   
   only some of its own parts but some of the parts of other machines;   
      
   It seems clear that an ensemble of greater than   
      
   n   
      
   Machines would have at least the potential for self reproduction so a   
   problem to be considered would be minimizing the size of the ensemble that   
   is coming as close as possible to some actual   
      
   n or 2n.   
      
   I mentioned two in here because fundamentally the work envelope problem   
   which is well discussed in the literature of mechanical engineering limits   
   progress toward minimizing the size of the ensemble. However machines such   
   as the bandsaw or the central is grinder have open work envelopes which   
   lend toward a resolution of the envelope problem since their work envelopes   
   are effectively infinite and practically larger than the size of any   
   conceivable part of any conceivable machine in the ensemble.   
      
      
   If we represent what we know in terms of a matrix connecting the elements   
   of the vector that we presume to exist, each to the other, cascading   
   towards the 1st element, that matrix is clearly a lower diagonal matrix,   
   and the problem turns out to be solvable...   
      
   Leaving as we say the proof for the interested reader we find:   
      
   d = o * ( I - Q ) ^ (-1)   
      
   Where the inverse represented turns out to be equal to the sum of all   
   powers of Q up to and including that power of Q in which the exponent is   
   equal to the rank of the matrix;   
      
   For that particular power proof can readily be established that the power   
   is equal to zero and so are many of the higher powers...   
      
   Other methods of finding the inverse exist...   
      
   However, if we multiply by the first power firstly we see that   
      
   d.1 = (o * Q).1   
      
   Represents what is called the first, first-order requirements.   
      
   This is a good thing because there always has to be a place to start with a   
   large project...   
      
   So with d.1 representing the first order requirements for an initial order   
   for a self-reproducing factory,   
      
   And because the d vector is in technological order,   
      
   d.1 is clearly the requirement first to be addressed;   
      
   So we can establish a status vector   
      
   s   
      
    the same length as   
      
   d...   
      
      
   Let's say that the value of d.1happens to be six or something like that;   
      
   Six nuts which fit six bolts later or something like that;   
      
   Or six flats which you have to be made from steel to make even one nut;   
      
   Or six positions in the jig which holds the blank for milling;   
      
   Or the six divisions of the unit circle represented by the imaginary number=   
   ,   
      
   i   
      
   to the sixth power;   
      
   Or maybe six minutes of machine time or something like that which we've   
   calculated beforehand by experiment and understanding, is how much time you   
   need to make some thing which we'll need later;   
      
   And let's say we've fulfilled the requirement so we have six of these   
   things on hand:   
      
   s.1<== d.1   
      
   And clearly, at this first success,   
      
   The rest of the vector   
      
   s   
      
    is equal to 0.   
      
   I use mathcad 6.0 Plus--yes that's the software I like for that...   
      
   It doesn't matter much what the difference between s and d actually is but   
   clearly when this difference is zero the project is a success....   
      
   So having defined success at least adequately,   
      
      
   Please look at the open collective under the name   
      
   The Replikon Net   
      
   And join me tomorrow downtown in Washington DC,   
      
   If you would like to....   
      
      
   Other than that,   
      
      
      
   Cheers, from Douglas MIT 1981 XXV NG   
      
   My handle is Godot.   
      
   If selected I will serve.   
      
      
      
   *Thursday, March 7 *   
   Hamilton Hotel   
   1001 14th St. NW, Washington, DC 20005   
      
   Copyright =C2=A9 Douglas Dana Goncz 2024   
      
   [[Mod. note -- 29 lines of legal notices snipped here.  -- jt]]   
      
   --- SoupGate-Win32 v1.05   
    * Origin: you cannot sedate... all the things you hate (1:229/2)