Ben, 

 

You're right.  Although one might seem to be getting a free lunch in terms
of being able to create more assemblies than the number of nodes from which
they are created, it would appear that the extra number of links required
not only for auto-associative activation withn an assembly, but that would
be required to activate an assembly from the outside with a signal that
would be distinguishable over the cross talk, may prevent such a use of node
assemblies from resulting in any actual saving.

 

If Vlad's forumula for a lower bound is correct, the one that I used in the
Excel spreadsheet I sent out earlier under this thread, then it is clear one
can create substantially more assemblies than nodes, with maximum overlaps
below 5%, but it is not clear the increased costs in extra connections would
be worth it, since it is not clear that the cost of a node, need be that
much higher than the cost of a link.

 

Ed Porter

 

-----Original Message-----
From: Ben Goertzel [mailto:[EMAIL PROTECTED] 
Sent: Tuesday, October 21, 2008 11:28 AM
To: agi@v2.listbox.com
Subject: Re: [agi] Who is smart enough to answer this question?

 


makes sense, yep...

i guess my intuition is that there are obviously a huge number of
assemblies, so that the number of assemblies is not the hard part, the hard
part lies in the weights...

On Tue, Oct 21, 2008 at 11:18 AM, Ed Porter <[EMAIL PROTECTED]> wrote:

Ben,

 

In my email starting this thread on 10/15/08 7:41pm I pointed out that a
more sophisticated version of the algorithm would have to take connection
weights into account in determining cross talk, as you have suggested below.
But I asked for the answer to a more simple version of the problem, since
that might prove difficult enough, and since I was just trying to get some
rough feeling for whether or not node assemblies might offer substantial
gains in possible representational capability, before delving more deeply
into the subject.

 

Ed Porter

 

 

-----Original Message-----
From: Ben Goertzel [mailto:[EMAIL PROTECTED] 

Sent: Monday, October 20, 2008 10:52 PM
To: agi@v2.listbox.com
Subject: Re: [agi] Who is smart enough to answer this question?

 


But, suppose you have two assemblies A and B, which have nA and nB neurons
respectively, and which overlap in O neurons...

It seems that the system's capability to distinguish A from B is going to
depend on the specific **weight matrix** of the synapses inside the
assemblies A and B, not just on the numbers nA, nB and O.

And this weight matrix depends on the statistical properties of the memories
being remembered.

So, these counting arguments you're trying to do are only going to give you
a very crude indication, anyway, right? 

ben



On Mon, Oct 20, 2008 at 5:09 PM, Ed Porter <[EMAIL PROTECTED]> wrote:

Ben, 

 

I am interested in exactly the case where individual nodes partake in
multiple attractors,  

 

I use the notation A(N,O,S) which is similar to the A(n,d,w) formula of
constant weight codes, except as Vlad says you would plug my varaiables into
the constant weight formula buy using A(N, 2*(S-0+1),S).

 

I have asked my question assuming each node assembly has the same size S for
to make the math easier.  Each such assembly is an autoassociative
attractor.  I want to keep the overlap O low to reduce the cross talk
between attractors.  So the question is how many node assemblies A, can you
make having a size S, and no more than an overlap O, given N nodes.

 

Actually the cross talk between auto associative patterns becomes an even
bigger problem if there are many attractors being activated at once (such as
hundreds of them), but if the signaling driving different the population of
different attractors could have different timing or timing patterns, and if
the auto associatively was sensitive to such timing, this problem could be
greatly reduced.

 

Ed Porter

 

-----Original Message-----
From: Ben Goertzel [mailto:[EMAIL PROTECTED] 

Sent: Monday, October 20, 2008 4:16 PM
To: agi@v2.listbox.com
Subject: Re: [agi] Who is smart enough to answer this question?

 


Wait, now I'm confused.

I think I misunderstood your question.

Bounded-weight codes correspond to the case where the assemblies themselves
can have n or fewer neurons, rather than exactly n.

Constant-weight codes correspond to assemblies with exactly n neurons.

A complication btw is that an assembly can hold multiple memories in
multiple attractors.  For instance using Storkey's palimpsest model a
completely connected assembly with n neurons can hold about .25n attractors,
where each attractor has around .5n neurons switched on.

In a constant-weight code, I believe the numbers estimated tell you the
number of sets where the Hamming distance is greater than or equal to d.
The idea in coding is that the code strings denoting distinct messages
should not be closer to each other than d.

But I'm not sure I'm following your notation exactly.

ben g

On Mon, Oct 20, 2008 at 3:19 PM, Ben Goertzel <[EMAIL PROTECTED]> wrote:

 

I also don't understand whether A(n,d,w) is the number of sets where the
hamming distance is exactly d (as it would seem from the text of
<http://en.wikipedia.org/wiki/Constant-weight_code>
http://en.wikipedia.org/wiki/Constant-weight_code ), or whether it is the
number of set where the hamming distance is d or less.  If the former case
is true then the lower bounds given in the tables would actually be lower
than the actual lower bounds for the question I asked, which would
correspond to all cases where the hamming distance is d or less.



The case where the Hamming distance is d or less corresponds to a
bounded-weight code rather than a constant-weight code.

I already forwarded you a link to a paper on bounded-weight codes, which are
also combinatorially intractable and have been studied only via
computational analysis.

-- Ben G

 




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CEO, Novamente LLC and Biomind LLC
Director of Research, SIAI
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Director of Research, SIAI
[EMAIL PROTECTED]

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CEO, Novamente LLC and Biomind LLC
Director of Research, SIAI
[EMAIL PROTECTED]

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overcome "  - Dr Samuel Johnson



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