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On 30 Jul 2005 10:55:40 -0700, "asdf" <qjohnny2000@xxxxxxxxx> wrote:
>Okay here is a example
>Square A ( or Block A )
>1 2
>3 4
>Square B ( or Block B )
>2 3
>4 6
>
>d( A, B ) = 7 = ( 1-2)^2 + (2-3)^2 + (3-4)^2 + (4-6)^2
>
>now replace each entry with the means.
>New A
>2.5 2.5
>2.5 2.5
>
>New B
>3.75 3.75
>3.75 3.75
>
>distance is 4*(3.75-2.5)^2 = 6.25
>
>Calling these things squares is not the best way of wording things !
Yes, better to call them points in R^n.
So in your example, if we view the "squares" as points in R^4, we can
write:
A=(1,2,3,4)
B=(2,3,4,6)
A'=(5/2,5/2,5/2,5,2)
B'=(15/4,15/4,15/4,15/4)
and the distances being compared are d(A,B), d(A',B'). The standard
distance in R^n is given by:
d(x,y)=sqrt(sum (x[i]^2-y^[i]^2, i=1..n))
but comparing distances squared will give the same order. Therefore
In the example above, the distance squared from A to B is 7 while the
distance squared from A' to B' is 6.25.
So the general question is:
If A,B are any arbitrary points of R^n, must d(A,B)>=d(A',B')?
The answer is yes. Here is a sketch of the proof:
Interpret A,B, A',B' as vectors.
Let u be a unit vector in R^n in the direction of (1,1,...1).
Note that A'=proj_u(A) and B'=proj_u(B) so
A'-B'=proj_u(A)-proj_u(B)=proj_u(A-B)
Then:
d(A,B)=|A-B|>=|proj_u(A-B)|=|A'-B'|=d(A',B')
Therefore d(A,B)>=d(A',B')
quasi
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