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Re: [WM]: Normalized Correlation
I wish that I had thought of this all by myself.
In fact this is just a part of the theorem that proves
that m-sequences are almost orthogonal. (ref. Pickholtz)
Cross correlation of m-sequences is like adding
using exclusive-or every integer from 1 to (2^N)-1 inclusive. There is
always only one more odd than even - which means that it gets close to
zero (giving a slightly negative cross correlation)
To make m-sequences orthogonal the missing all zero
state needs to be included. This is not hard to do in
software but in hardware a shift register would
lock in an all zero state.
Joe
Cagatay Candan wrote:
>
> At 10:50 AM 8/14/02 -0400, Joseph O'Ruanaidh wrote:
> >However, given that these are BITMAP images we might want to try
> HAMMING
> >distance. This would be the number of bits that are different. The
> >maximum number of bits they can have in common is N, the length of
> >the
>
> >vector, so let's divide by N to normalize. Okay that works.
> >
> >What is the relationship between the two measures?
> >The first is based on Euclidean distance (sum of squares)
> >
> >Look at these tables:
> >
> >XOR 1 0
> >----------------
> >1 0 1
> >0 1 0
> >
> >
> >x -1 1
> >----------------
> >-1 1 -1
> >1 -1 1
> >I hope you can see that they are isomorphic. If you
> >changed all your binary zeros to ones and binary ones to minus ones
> >the
>
> >two distances would be equivalent (different numbers saying exactly
> >the
>
> >same thing).
>
> This is a great observation. To see the isomorphism more clearly,
> please have a look at this table:
>
> set_1= { even, odd } op_1={addition}
> set_2= { negative, positive} op_2={multiplication}
>
> set_1& op_1 % set_2 & op_2
> %%%%%%%%%%%%%%%%%%%%%%%%%%
> even+even= even % positive*positive=positive
> even+odd = odd % positive*negative=negative
> odd+even = odd % negative*positive=negative
> odd+odd = even % negative*negative=positive
> %%%%%%%%%%%%%%%%%%%%%%%%%%
>
> The essence of the whole thing is: Both operations map the set
> elements identically. Therefore I may call this operation as binary
> addition if that is my wish. And my marsian friend may call it
> something else if that is
>
> his wish. But his computer and my computer are doing **exact** same
> thing when we execute this routine. Just we have different labels on
> it.
>
> best wishes,
> cagatay
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