SDM_the_search_for_powers_of_2_everywhere.tex

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    \title{SDM\_the\_search\_for\_powers\_of\_2\_everywhere}




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    \begin{document}


    \maketitle




    \section{Sparse Distributed Memory: How many
subspaces?}\label{sparse-distributed-memory-how-many-subspaces}

In his original analysis, Kanerva uses... ...1,000 dimensions
...1,000,000 circles ...with radius=451 bits so that a random bistring
will be stored in aproximately 1,000 hard locations.

A first observation here is that powers of 10 are not well suited,
either for the mathematical analysis of the space, or for computer
science. Let us play a little with numbers here, starting from Kanerva's
parameters. Our goal will be to find how many circles we should use and
the radius of a circle. Can we find some optimum set of parameters?

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number of dimensions= 1000

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{\color{incolor}In [{\color{incolor}5}]:} \PY{k}{def} \PY{n+nf}{analysis}\PY{p}{(}\PY{n}{n}\PY{p}{)}\PY{p}{:}
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            \PY{n+nb}{print} \PY{p}{(}\PY{l+s+s1}{\PYZsq{}}\PY{l+s+s1}{orthoghonal distance=}\PY{l+s+s1}{\PYZsq{}}\PY{p}{,}\PY{n}{mu}\PY{p}{(}\PY{n}{n}\PY{p}{)}\PY{p}{)}
            \PY{n+nb}{print} \PY{p}{(}\PY{l+s+s1}{\PYZsq{}}\PY{l+s+s1}{standard deviation=}\PY{l+s+s1}{\PYZsq{}}\PY{p}{,}\PY{n}{sigma}\PY{p}{(}\PY{n}{n}\PY{p}{)}\PY{p}{)}
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            \PY{n+nb}{print} \PY{p}{(}\PY{l+s+s1}{\PYZsq{}}\PY{l+s+s1}{3 sigma=}\PY{l+s+s1}{\PYZsq{}}\PY{p}{,} \PY{l+m+mf}{3.09}\PY{o}{*}\PY{n}{sigma}\PY{p}{(}\PY{n}{n}\PY{p}{)}\PY{p}{)}
            \PY{n+nb}{print} \PY{p}{(}\PY{l+s+s1}{\PYZsq{}}\PY{l+s+s1}{the distance from a pole to the equator=}\PY{l+s+s1}{\PYZsq{}}\PY{p}{,} \PY{n}{math}\PY{o}{.}\PY{n}{sqrt}\PY{p}{(}\PY{n}{n}\PY{p}{)}\PY{p}{,} \PY{l+s+s1}{\PYZsq{}}\PY{l+s+s1}{standard deviations}\PY{l+s+s1}{\PYZsq{}}\PY{p}{)}
            \PY{n+nb}{print} \PY{p}{(}\PY{l+s+s1}{\PYZsq{}}\PY{l+s+s1}{the percentage of n (in relation to the range R Bits) is}\PY{l+s+s1}{\PYZsq{}}\PY{p}{,} \PY{n}{percentage\PYZus{}of\PYZus{}n}\PY{p}{(}\PY{n}{n}\PY{p}{)}\PY{p}{)}\PY{c+c1}{\PYZsh{} When using mu \PYZhy{} 3 sigma...}
            \PY{n+nb}{print} \PY{p}{(}\PY{l+s+s1}{\PYZsq{}}\PY{l+s+s1}{Because 3 standard deviations contain \PYZti{}1/1000 of a normal distribution, }\PY{l+s+s1}{\PYZsq{}} \PYZbs{}
                    \PY{l+s+s1}{\PYZsq{}}\PY{l+s+s1}{approximately one in a thousand items will be found in a circle of}\PY{l+s+s1}{\PYZsq{}}\PY{p}{,} \PYZbs{}
                    \PY{n}{mu}\PY{p}{(}\PY{n}{n}\PY{p}{)}\PY{o}{\PYZhy{}}\PY{l+m+mf}{3.09}\PY{o}{*}\PY{n}{sigma}\PY{p}{(}\PY{n}{n}\PY{p}{)}\PY{p}{,}\PY{l+s+s1}{\PYZsq{}}\PY{l+s+s1}{ radius}\PY{l+s+s1}{\PYZsq{}}\PY{p}{)}
            \PY{n+nb}{print} \PY{p}{(}\PY{l+s+s1}{\PYZsq{}}\PY{l+s+s1}{...........................................}\PY{l+s+se}{\PYZbs{}n}\PY{l+s+se}{\PYZbs{}n}\PY{l+s+se}{\PYZbs{}n}\PY{l+s+se}{\PYZbs{}n}\PY{l+s+s1}{\PYZsq{}}\PY{p}{)}
\end{Verbatim}

    \begin{Verbatim}[commandchars=\\\{\}]
{\color{incolor}In [{\color{incolor}6}]:} \PY{k}{for} \PY{n}{n} \PY{o+ow}{in} \PY{p}{[}\PY{l+m+mi}{100}\PY{p}{,} \PY{l+m+mi}{1000}\PY{p}{,} \PY{l+m+mi}{10000}\PY{p}{,} \PY{l+m+mi}{256}\PY{p}{,} \PY{l+m+mi}{1024}\PY{p}{]}\PY{p}{:}
            \PY{n}{analysis}\PY{p}{(}\PY{n}{n}\PY{p}{)}
\end{Verbatim}

    \begin{Verbatim}[commandchars=\\\{\}]
************ ANALYSIS FOR n= 100
orthoghonal distance= 50.0
standard deviation= 5.0
the space offers  20.0  standard deviations.
3 sigma= 15.45
the distance from a pole to the equator= 10.0 standard deviations
the percentage of n (in relation to the range R Bits) is 30.0
Because 3 standard deviations contain \textasciitilde{}1/1000 of a normal distribution, approximately one in a thousand items will be found in a circle of 34.55  radius
{\ldots}




************ ANALYSIS FOR n= 1000
orthoghonal distance= 500.0
standard deviation= 15.811388300841896
the space offers  63.0  standard deviations.
3 sigma= 48.857189849601454
the distance from a pole to the equator= 31.622776601683793 standard deviations
the percentage of n (in relation to the range R Bits) is 9.486832980505138
Because 3 standard deviations contain \textasciitilde{}1/1000 of a normal distribution, approximately one in a thousand items will be found in a circle of 451.14281015039853  radius
{\ldots}




************ ANALYSIS FOR n= 10000
orthoghonal distance= 5000.0
standard deviation= 50.0
the space offers  200.0  standard deviations.
3 sigma= 154.5
the distance from a pole to the equator= 100.0 standard deviations
the percentage of n (in relation to the range R Bits) is 3.0
Because 3 standard deviations contain \textasciitilde{}1/1000 of a normal distribution, approximately one in a thousand items will be found in a circle of 4845.5  radius
{\ldots}




************ ANALYSIS FOR n= 256
orthoghonal distance= 128.0
standard deviation= 8.0
the space offers  32.0  standard deviations.
3 sigma= 24.72
the distance from a pole to the equator= 16.0 standard deviations
the percentage of n (in relation to the range R Bits) is 18.75
Because 3 standard deviations contain \textasciitilde{}1/1000 of a normal distribution, approximately one in a thousand items will be found in a circle of 103.28  radius
{\ldots}




************ ANALYSIS FOR n= 1024
orthoghonal distance= 512.0
standard deviation= 16.0
the space offers  64.0  standard deviations.
3 sigma= 49.44
the distance from a pole to the equator= 32.0 standard deviations
the percentage of n (in relation to the range R Bits) is 9.375
Because 3 standard deviations contain \textasciitilde{}1/1000 of a normal distribution, approximately one in a thousand items will be found in a circle of 462.56  radius
{\ldots}





    \end{Verbatim}

    \begin{Verbatim}[commandchars=\\\{\}]
{\color{incolor}In [{\color{incolor}86}]:} \PY{n}{print\PYZus{}analysis}\PY{p}{(}\PY{l+m+mi}{1024}\PY{p}{)}
\end{Verbatim}

    \begin{Verbatim}[commandchars=\\\{\}]
************ ANALYSIS FOR n= 1024
orthoghonal distance= 512.0
standard deviation= 16.0
3 sigma= 48.0
the distance from a pole to the equator= 32.0 standard deviations
the percentage of n (in relation to the range R Bits) is 9.375
Because 3 standard deviations contain \textasciitilde{}1/1000 of a normal distribution, approximately one in a thousand items will be found in a circle of 464.0  radius
\#\#\#\#\#\#\#\#\#\#\#\#\#\#\#\#\#\#\#\#\#\#\#\#\#\#\#\#\#\#\#\#\#\#\#\#\#\#\#\#\#\#\#\#\#\#\#\#\#\#\#\#\#\#\#\#\#\#\#\#\#\#\#

    \end{Verbatim}

    Let us focus on the word \emph{aproximately}. The estimation is given by
usual tables with metrics of the normal curve.

The \emph{unit} of analysis provided is that of the standard deviation.

Our intention is to make the bit the unit of analysis, and estimate the
statistics of the system in a more precise form.

    \begin{Verbatim}[commandchars=\\\{\}]
{\color{incolor}In [{\color{incolor} }]:}
\end{Verbatim}


    % Add a bibliography block to the postdoc



    \end{document}