Newsgroups: rec.puzzles
From: haoyuep@aol.com (Dan Hoey)
Date: 28 Oct 2003 03:29:33 GMT
Subject: Re: R = PI ohms

I wrote:
> Less contrived would be "no resistors can have their ends
> wired together" (along with the expected "both ends of each
> resistor must be connected to both terminals through a path
> not including that resistor").

and Phil Carmody correctly observes that this still allows us
to throw away a pair of resistors wired in parallel to an
isolated node.  I now believe the right criterion is for the
graph of resistances (plus an edge for the source-to-sink
circuitry) to be _biconnected_, which means that the removal
of any one node will not result in a disconnected graph.
This implies all the previous criteria except the bogus one
that prohibits a resistor that fails to pass any current due
to the values of the resistances rather than the topology of
the graph.

In addition, I should mention that I've re-derived the star-
clique rule, which allows us to remove any internal node of
the graph.  If internal node  x  is connected by conductances
y_i  to nodes  n_i,  we may remove  x  and the  y_i  and
replace them with conductances  y_ij = y_j y_k / (sum_i y_i)
between each pair of nodes  n_i, n_j.

Proof:  Suppose in the original network that nodes  n_i  are
at voltages  v_i,  and  x  is at voltage  v.  By Kirkhoff's
law,

  0 = sum_i (v - v_i) y_i
    = (v sum_i y_i) - (sum_i v_i y_i),  so

  v = (sum_i v_i y_i)/(sum_i y_i).

The current out of node  n_k  will be  y_k (v_k - v).

In the proposed network, if the nodes  n_i  are at voltages
v_i,  then the current out of  n_k  will be

    sum_j  (v_k - v_j) y_jk
  = sum_j  (v_k - v_j) y_j y_k / (sum_i y_i)
  = y_k  (sum_j  (v_k - v_j) y_j)  /  (sum_i y_i)
  = y_k (  (sum_j v_k y_j)/(sum_i y_i)
         - (sum_j v_j y_j)/(sum_i y_i) )
  = y_k (v_k - v),

which is the same as the current out of  n_k  in the original
network.  By uniqueness of the solution to any network, this
implies that the voltages and currents will be preserved in
the resulting network,  QED.

This allows us to remove the internal nodes one by one, which
should be reasonably efficient if we collapse parallel
conductances after each removal.  The resemblance of this
procedure to Gaussian elimination reminds me of that of
Persian Monarchs to Blind Hookey.

Dan Hoey
haoyuep@aol.com