Re: Need help on application of the normalization formula forCSE494

[Subbarao Kambhampati <rao@asu.edu>]

From: SHAKIL ZERO <Shakil.Zero@asu.edu>
Subject: Need help on application of the normalization formula for CSE494
Date: Sat, 12 Oct 2002 01:34:00 -0400
Message-ID: <LIAPDGECMFMGHAAA@mailcity.com>

Shakil.Zero> Dear Dr. Rao,
Shakil.Zero> 
Shakil.Zero> I would appreciate it,if you could please explain how the
Shakil.Zero> normalization formula works for the Hubs and the
Shakil.Zero> Authority values calculations. The slide, Authority and
Shakil.Zero> Hub Pages (9), gives the formula but the application of
Shakil.Zero> that doesn't make sense.  As an example, will you be able
Shakil.Zero> to send me the process that how the normalization of
Shakil.Zero> a(q1) was calculated to 0.267 on the Slide, Authority and
Shakil.Zero> Hub Pages (11)?

Sure. 
The authority vector is a 5-tuple

<1 0 0 3 2>

To normalize the vector you simply divide each element by the
magnitude.

Magnitude is sqrt(sum square of all elements)

sqrt(1+0+0+9+4)=sqrt(14)=3.741

So normalized vector is 

1/3.741* <1 0 0 3 2>

Thus, normalized value of a(q1) is 1/3.741 which is 0.267

a(q4) will be 3/3.741=0.802 and a(q5)=2/3.741=0.535

Rao


Shakil.Zero> 
Shakil.Zero> Thanks,
Shakil.Zero> 
Shakil.Zero> Shakil
Shakil.Zero> 
Shakil.Zero> 
Shakil.Zero> 
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