Re: View updating in practice?
Date: Wed, 13 Nov 2002 06:36:09 GMT
Message-ID: <tpmA9.13340$QZ.4692_at_sccrnsc02>
"Jens Lechtenbörger" <lechtej_at_uni-muenster.de> wrote in message
news:m265v6nz31.fsf_at_pcwi1068.uni-muenster.de...
Let's doublecheck that I understand your approach. I'm going to use
a very restricted set of views -- Linear Algebra
Transformations only. (Then, we need numbers domain with operators,
> Just to clarify the connection between view inverses and constant
> complement translators, here is a rough outline of my approach:
select x+2*y+3*z from D
select 2*x+y+0*z from D
Linear transformation in general could map a set of linear independent
vectors into a set of linear independent vectors. Then, it preserves
all the "information" contained in that linear set. Otherwise, it
preserves information "partially". In my example, linear independent set of
vectors {<x=1, y=0, z=0>,<x=0, y=1, z=0>} map into linear independent
set {<1,2>,<2,1>}. If we add one more linear independent vector, say,
<x=0, y=0, z=1>, then the result {<1,2>,<2,1>,<3,0>} is no longer
linear independent.
> Now, a set C of
> views over D is a complement of V if it is possible to compute D
> from V and C, i.e., if there is way to compute any source state from
> the state of V and C,
> Note that in
> general, complements are not unique (as you might expect from
> algebraic complements).
> In particular, views that describe a copy
> of D are a complement for any set of views.
View defined as a "Copy of D" is just a view based upon identity matrix:
select x, y, z from D
> Now, a view update on view V can be translated under constant
> complement if there is a complement C of V such that the contents of
> C do not change when the view update is executed on the database.
Here I'm lost.
Suppose, D current content is {<x=1,y=1,z=1>}. A view V would manifest
select x-z from D
It would depend on the new tuple and the D currrent content, of course.
Constant complement, however, seems meaningless from linear algebra
perspective: neither user, nor the systems calculates complement
view. Given a view update, one needs to translate it back to the
base relations. I don't see how can it be done without solving
<1*1+2*1+3*1,2*1+1*1>=<6,3>. Suppose a new tuple
<x=2,y=2,z=2> added. A view V would manifest an extra tuple
<1*2+2*2+3*2,2*2+1*2>=<12,6>. Is there a view D such that an image
of the new and the old tuple is the same? It certainly is, for example
Or maybe I'm just missing correct Linear Algebra interpretation of the constant complement? Received on Wed Nov 13 2002 - 07:36:09 CET