# Re: Counting propositions

From: x <x-false_at_yahoo.com>
Date: Tue, 22 Jun 2004 22:51:12 +0300
Message-ID: <40d88cae\$1_at_post.usenet.com>

"Mikito Harakiri" <mikharakiri_at_iahu.com> wrote in message news:h60Cc.24\$da4.292_at_news.oracle.com...
>
> "x" <x-false_at_yahoo.com> wrote in message
news:40d884f8\$1_at_post.usenet.com...
> > > Let's go through the list:
> > >
> > > AVG - nonfundamental, can be expressed in terms of SUM
> > > STDDEV = SQRT(VARIANCE)
> > > VARIANCE = (SUM(x*x)-(SUM(x)*SUM(x))/SUM(1))/(SUM(1)-1)
> > >
> > > Do you need mere examples to be convinced that there are only 4
> > fundamental
> > > aggregate operators?
> > > 1. SUM = "+"*
> > > 2. MIN = "/\"*
> > > 3. MAX = "\/"*
> > > 4. LIST = "||"*
> >
> > Nonsense.
> > SUM([1])=1.
> > COUNT([1,2,3,4])=4 !=SUM([1,2,3,4]).
>
> SUM(1) is abbreviation for "select SUM(1) from T"
>
> In other words "1" in SQL is a function that maps column value x into 1
for
> all x.
>
> In order to calculate "Count([1,2,3,4])"
> (where I deliberately changed the case in order to avoid confusion with
SQL
> syntax) we first apply function f:x->1 to each element of the bag.
> Therefore,

> Count([1,2,3,4]) = Sum([f(1),f(2),f(3),f(4)])=Sum([1,1,1,1])
Yes. And for every such function "f" you get some more aggregate operators ...

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Received on Tue Jun 22 2004 - 21:51:12 CEST

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