Re: Extending my question. Was: The relational model and relational algebra - why did SQL become the industry standard?
Date: Thu, 27 Feb 2003 10:45:40 -0000
Message-ID: <b3kqqg$sqa$1_at_sp15at20.hursley.ibm.com>
"Steve Kass" <skass_at_drew.edu> wrote in message
news:b3k8ks$kca$1_at_slb5.atl.mindspring.net...
[snpi]
> I'm lost on who is countering whose argument here, but you have
> described one type that represents the 9 points on and at the
> center of the 2 x 2 square at the origin of Z x Z, and another type
> that represents 9 points on and at the center of a circle (I think you
> may have meant to allow -0.75PI, ..., 0.75PI, 1 PI), instead of twice
> these values).
>
> These certainly represent very different subsets of cartesian space,
> so what do you mean by saying these are equivalent? Once you stray
> from points of the cartesian form (N,0) and (0,N), I don't think you'll
> have much luck finding equivalent polar and cartesian representations.
I was attempting to counter my own argument. With a flawed example as you
point out.
However if I fix the example, I think I do have equivalent polar and cartesian
representations outside of (N,0) and (0,N).
To support these 9 points
( 0, 0) = ( 0, 0) ( 1, 0) = ( 0, 1) ( 1, 1) = ( 0.25PI, 1SQRT2) ( 0, 1) = ( 0.5PI, 1) (-1, 1) = ( 0.75PI, 1SQRT2) (-1, 0) = ( -PI, 1) (-1, -1) = (-0.75PI, 1SQRT2) ( 0, -1) = ( -0.5PI, 1) ( 1, -1) = (-0.25PI, 1SQRT2)
we could have a type such as
TYPE TINY_INT POSREP( {-1,0,1 } );
TYPE TINY_PI
POSREP( {-0.75 PI, -0.5 PI, -0.25 PI, 0, 0.25 PI, 0.5 PI, .75} );
TYPE TINY_R2 POSREP( { 0, 1SQRT2, 1 } );
TYPE TINY_POINT
POSREP POINT_XY (X TINY_INT, Y TINY_INT)
POSREP POINT_RT (T TINY_PI, R TINY_R2
CONSTRAINT IF R = 0 SQRT2 THEN T = 0 IF T IN (0.5PI, 0, -PI, -0.5PI) THEN R = 1 ELSE R = 1SQRT2 );
I guess things get a little more involved when X or Y > 2, but I'm happy for the moment.
Regards
Paul Vernon
Business Intelligence, IBM Global Services
Received on Thu Feb 27 2003 - 11:45:40 CET