Re: Using the RM for ADTs

From: Brian Selzer <brian_at_selzer-software.com>
Date: Thu, 9 Jul 2009 09:16:07 -0400
Message-ID: <s4m5m.9183\$Jb1.432_at_flpi144.ffdc.sbc.com>

"David BL" <davidbl_at_iinet.net.au> wrote in message news:48a4205b-e46f-4cb7-8c9e-d2a923135c80_at_x5g2000prf.googlegroups.com...
> On Jul 9, 10:11 am, "Brian Selzer" <br..._at_selzer-software.com> wrote:
>> "David BL" <davi..._at_iinet.net.au> wrote in message
>
>> > As an example, consider a circuit consisting of 12 x 1 ohm resistors
>> > and 8 nodes wired up in the manner of a 3-dimensional cube. All the
>> > resistors are indistinguishable and all the nodes are
>> > indistinguishable, even in the context of the circuit that they appear
>> > in.
>>
>> They are not indistinguishable. Just pick an arbitrary component lead,
>> and
>> the rest can be described in terms of it because each has different paths
>> to
>> it.
>
> Well, obviously some things won't be indistinguishable anymore after
> you "just pick" one. Picking one means distinguishing it from all
> others.
>
> Note that distinguishing one of the nodes is insufficient to break all
> symmetry. Call it "top". This immediately distinguishes a node on
> the opposite corner (call it "bottom"). However there remains 3-way
> symmetry in the nodes adjacent to "top", and a further 2-way symmetry
> from any one of these nodes to the next three adjacent nodes as we
> proceed towards the bottom of this lattice structure.
>
> In your case picking a component lead can be regarded as picking an
> ordered pair of adjacent nodes. That still leaves a 2-way symmetry as
> described above.

I wasn't denying that there is symmetry, nor that it poses problems for identification. What I'm arguing is that despite the symmetry, there are still 8 nodes and 12 resistors in the cube example, which means that there must be a means to distinguish between the nodes and the resistors, for things that are indistinguishable are the same thing: that is the essence of identity. The nodes and components in a circuit template can be thought of as verticies and hyperedges in a connected hypergraph. Certainly each vertex can be distinguished from all other verticies and each hyperedge can be distinguished from all other hyperedges in the same hypergraph, can't they? Received on Thu Jul 09 2009 - 15:16:07 CEST

Original text of this message