Hi Jeff,

That is what the spatial pooler does. See pages 21-22 of the white paper
for more details. Each column in an HTM region receives input from a unique
subset of input bits, so when the region receives a pattern of input bits,
regardless of what percentage of the input bits are on, some columns will
receive more excitation than others. Only the 2% of columns that receive
the highest amount of excitation will be activated. So regardless of what
proportion of input bits are active, the region will end up with a sparse
distributed representation of the input, with 2% of columns active.

-Mike

_____________
Michael Ferrier
Department of Cognitive, Linguistic and Psychological Sciences, Brown
University
[email protected]


On Sun, Oct 20, 2013 at 9:40 PM, Jeff Fohl <[email protected]> wrote:

> Hello -
>
> I hope this is not being posted to the wrong list. This is my first post
> here. Please let me know if there is a more appropriate place for this
> question.
>
> In preparation for learning NuPIC, I have read "On Intelligence", and I am
> now reading the HTM white paper put out by Numenta.
>
> Making my way through the white paper, I got stuck on one passage, which I
> can't really make sense of. Wondering if anyone can help me through this
> part. The passage in question is on pages 11-12 of the white paper PDF -
> specifically the second paragraph included below.
>
> *HTM regions also use sparse distributed representations. In fact, the
> memory mechanisms within an HTM region are dependent on using sparse
> distributed representations, and wouldn’t work otherwise. The input to an
> HTM region is always a distributed representation, but it may not be
> sparse, so the first thing an HTM region does is to convert its input into
> a sparse distributed representation.*
>
> *For example, a region might receive 20,000 input bits. The percentage of
> input bits that are “1” and “0” might vary significantly over time. One
> time there might be 5,000 “1” bits and another time there might be 9,000
> “1” bits. The HTM region could convert this input into an internal
> representation of 10,000 bits of which 2%, or 200, are active at once,
> regardless of how many of the input bits are “1”. As the input to the HTM
> region varies over time, the internal representation also will change, but
> there always will be about 200 bits out of 10,000 active. *
>
> So, what exactly is going on here? How does a fluctuating input flow of
> 20,000 bits get converted into 200 bits? Obviously there is something
> important going on here, but I don't understand what it is. Any help
> illuminating this would be greatly appreciated!
>
> Many thanks,
>
> Jeff
>
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