From - Sat Jul  5 18:35:09 1997
From: momo@sbphy.physics.ucsb.edu (Monwhea Jeng)
Newsgroups: sci.physics.electromag,sci.electronics,rec.puzzles,sci.math
Subject: Re: Seven Resistor Puzzles
Date: 3 Jul 97 19:51:26 GMT
Organization: University of California, Santa Barbara

elkies@ramanujan.harvard.edu (Noam Elkies) writes:

>In article <5pcjt6$80t@clarknet.clark.net>, Keith Lynch <kfl@clark.net> wrote:
>>Puzzle 7:  Is there any arrangement of resistors which can give a
>>transcendental total resistance?

>Puzzle 8: Show that *any* positive real number can be obtained
>as the total resistance of an arrangement of unit resistors, and
>that this arrangement may be realized in the plane (without
>crossings).

SPOILER SPACE :





















Hey, neat puzzles.

Given unit resistors, I can make circuits with resistance
k and 1/k for all integer k>1, by hooking the resistors up in series 
or parallel, respectively. So for the rest of the problem assume
that I have resistors of this type as well. A resistance r>0
can be written in continued fraction form :

         1     1     1     1
r= k0 + --- + --- + --- + --- + . . .
        k1+   k2+   k3+   k4+

where the ki's are all integers, and it is understood that each
fraction is nested in the one before it. i.e.
r=k0+1/(k1+(1/(k2+1/(k3+1/(k4+. . .)))))

To see that we can do this, write r=k0+r0, with k0 the integer part of
r, and r0 the fractional part. r0=1/(1/r0), and (1/r0)>1,
so we can write (1/r0)=k1+r1, where k1>1 and is an integer, and
0<=r1<1. If r1=0 we stop writing the fraction. Otherwirse
0<r1<1, and we write r1=1/(1/r1)=1/(k2+r2). Next r2=1/(k3+r3),
r3=1/(k4+r4), . . . We keep on going, either eventually reaching
rn=0, and stopping, or going on forever. Convergence of this
process is left as an exercise to the reader.

We can make r by putting k0 and 
Now we see that we can make r by putting 1/k1 and 1/r1 in parallel.
We can make 1/r1 by putting k2 and r2 in series.
We can make r2 by putting 1/k3 and 1/r3 in parallel.
We can make 1/r3 by putting k4 and r4 in series.
We can make r4 by putting 1/k4 and 1/r4 in parallel.
And so on, forever. The arrangement in the plane is

 .---k0---.---k2---.---k4---.---k6---.---k8---   . . . .
          |        |        |        |           
         1/k1     1/k3     1/k5     1/k7 
          |        |        |        |
 .--------.--------.--------.--------.--------

where 1/k1 is short for k1 resistors in parallel.

I'm still working on the infinite plane problem. It seems
hard -- am I going to need to pull out the heavy mathematics
artillery or just be clever?

Momo