Newsgroups: rec.puzzles
From: haoyuep@aol.com (Dan Hoey)
Date: 26 Oct 2003 03:34:09 GMT
Subject: Re: R = PI ohms

"Martin Round" nos...@blueyonder.co.uk wrote:
> "Daniel W. Johnson" wrote:
> > A requirement to use all the resistors would not
> > actually be interesting.[...]

> True.  I suppose you could say, 'current must flow in each
> of the resistors', but that's a bit contrived.

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").

> Anyway, as expected, 10 resistors can do better than
> 9.  I'm not sure if what follows is the best solution,
> but it's the best one I've found yet.

I get 524935/167092, for an error of 1.95e-9.

> By the way, what method did you use to work out the exact
> resistance, expressed as two integers?  I've done it using
> simultaneous equations, but is there an easier way?
> Matrices perhaps?

Presumably you know the formulas for parallel and series
circuits.  Just one more formula is necessary for the
circuits we've seen so far: the "delta-Y" formula.  In any
circuit in which resistors of value  1/Yab, 1/Ybc, 1/Yca
appear between nodes  a and b,  b and c,  and  c and a,
respectively, we may remove those resistors and replace them
with a new node and resistors of value  1/Ya, 1/Yb, 1/Yc
between the new node and nodes  a, b, and c,  respectively,
where  Ya = (Yab Ybc  +  Ybc Yca  +  Yca Yab) / Ybc,  etc.
This transform can be inverted with the Y-delta transform
Yab = Ya Yb / (Ya + Yb + Yc), etc.

If you have a circuit without triangles, this won't help
you.  There is a star-clique transform that will turn any
circuit into a nest of resistors, but at this point it might
be as well to invoke a theorem of Kirkhoff:

Define the conductance of a subgraph to be the product of
the conductances (1/resistances) of its edges.  Then the
resistance of the network between nodes  A  and  B  is
the quotient of

  the sum of the conductances of all spanning trees of the network

divided by

  the sum of the conductances of all subgraphs that would
  become spanning trees if supplemented by an edge AB.

I have only seen this theorem quoted without proof, so I
can't answer a lot about it.  Still, this should be in
textbooks if you know where to look.

Dan Hoey
haoyuep@aol.com