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From: vaneric on 23 Jul 2008 01:38
hi
i am studying linear algebra and was reading thru some code processing
image data.i came across a piece of code that does some normalising as
below

public demoNorm(double[][] eigenfacesArray){
for(int i=0;i<eigenfacesArray.length;i++){
double norm=norm(eigenfacesArray[i]);
for(int x=0;x<eigenfacesArray[i].length;x++){
double v=eigenfacesArray[i][x];
eigenfacesArray[i][x]=v/norm;

}

}

}

private double norm(double[][] x){
int nr= x.length;
int nc= x[0].length;
double val=0.0;
for(int r= 0; r< nr; r++){
for(int c= 0; c< nc; c++){
val+= (x[r][c])* (x[r][c]) ;
}
}
return val;
}


when i tried this for a sample double[][] as below
double[][]=new double[][]{
{2.4,6.4,1.2,7.6,4.3},
{1.5,3.1,6.4,5.7,2.5},
{7.8,6.8,4.6,3.5,4.3},
{9.8,7.8,6.5,1.2,3.3}
};

i found that norm() yields a value equal to square of Frobenius norm
( ie ,norm()==> 586.410 while Frob norm=24.2159)

I am no expert in linear algebra,so i want to know if this norm has
any particular name.Also,is the above way of normalisation valid? or
do i have to find the sq.root of norm before i divide the elements of
the double[][]? can some experts help me with this?

thanks
eric
From: Eric Sosman on 23 Jul 2008 08:58
vaneric wrote:
> hi
> i am studying linear algebra and was reading thru some code processing
> image data.i came across a piece of code that does some normalising as
> below
>
> public demoNorm(double[][] eigenfacesArray){
> for(int i=0;i<eigenfacesArray.length;i++){
> double norm=norm(eigenfacesArray[i]);
> for(int x=0;x<eigenfacesArray[i].length;x++){
> double v=eigenfacesArray[i][x];
> eigenfacesArray[i][x]=v/norm;
>
> }
>
> }
>
> }
>
> private double norm(double[][] x){

Note that this is not the norm() method called by the
code above. This one takes a double[][] an argument, but
the earlier code uses a different method that takes a double[].

> int nr= x.length;
> int nc= x[0].length;
> double val=0.0;
> for(int r= 0; r< nr; r++){
> for(int c= 0; c< nc; c++){
> val+= (x[r][c])* (x[r][c]) ;
> }
> }
> return val;
> }
>
>
> when i tried this for a sample double[][] as below
> double[][]=new double[][]{

This won't compile.

> {2.4,6.4,1.2,7.6,4.3},
> {1.5,3.1,6.4,5.7,2.5},
> {7.8,6.8,4.6,3.5,4.3},
> {9.8,7.8,6.5,1.2,3.3}
> };
>
> i found that norm() yields a value equal to square of Frobenius norm
> ( ie ,norm()==> 586.410 while Frob norm=24.2159)
>
> I am no expert in linear algebra,so i want to know if this norm has
> any particular name.Also,is the above way of normalisation valid? or
> do i have to find the sq.root of norm before i divide the elements of
> the double[][]? can some experts help me with this?

How can I help you understand code you haven't shown?
Look at it this way: You're dealing with some body of code
which we'll call X. Instead of posting X and asking about it,
you instead post X', an inaccurate paraphrase of X. There's
not much I can do with X' as it stands -- as noted above, it
won't even compile -- so I change it to turn it into X'', my
guess about the nature of the hidden X. What are the chances
that X'' is the same as X? If I explain a fine point of X'',
will that help you understand what X is doing?

"Doctor, look at this X-ray of my foot and tell me why
I'm in such pain."

"Vaneric, this is an X-ray of a hand."

"Oh c'mon, Doc, hands and feet are evolutionary cousins,
not really that much different. Hurry up and get on with
the diagnosis, will you?"

--
Eric Sosman
esosman(a)ieee-dot-org.invalid
From: vaneric on 23 Jul 2008 14:12
On Jul 23, 5:58 pm, Eric Sosman <esos...(a)ieee-dot-org.invalid> wrote:
> How can I help you understand code you haven't shown?

sorry,
my mistake in copying the methods
there were 2 versions of norm(), one that takes in a double[][] and
another that takes in a double[]

for the demoNorm() above,the following method was used
private double norm(double[] x){
double val=0.0;
for(int i=0;i<x.length;i++){
val+=(x[i]*x[i]);
}
return val;
}


for my sample double[][] i used the other version (which takes in a
double[][])

sorry again for the silly mistake of a beginner
eric
p.s: great reply..really enjoyed it!





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