Re: Possible problems with Date & McGoveran View Updating

Mikito Harakiri wrote:
> "Jan Hidders" <jan.hidders_at_pandora.be> wrote in message
> news:Ym%8b.20114$7y3.1376399_at_phobos.telenet-ops.be...

>> Mikito Harakiri wrote:
>> > "Jan Hidders" <jan.hidders_at_pandora.be> wrote in message
>> > news:PX38b.14807$2Z.580673_at_phobos.telenet-ops.be...
>> >> A small example. Suppose we have a table Emps(name,dept) and a view
>> >> TrEmps defined by
>> >>
>> >>   SELECT name
>> >>   FROM Emps
>> >>   WHERE dept = "sales";
>> >>
>> >> then if you add a tuple to this view it is perfectly clear which tuple
>> >> should be added to Emps and so this is a well-defined update.
>> >
>> > This view is not updateable. If you delete a tuple from this view, do
>> > you
>> > delete from the base table or update dept = 'sales' into some other
>> > value
>> > (which we can't deduce from TrEmps alone)?
>>
>> According to the definition I gave this deletion is well-defined. The
>> unique
>> "minimal" solution is that the corresponding tuple is deleted from Emps,
>> and indeed that is all that the database knows for certain: this tuple
>> has to to be removed. I find this quite intuitive, can you explain why
>> you don't?

No, this is an illegal update because you are trying to insert a tuple into a relation that cannot have such a tuple. This makes sense because your view is your ontology so the only thing that you can tell the database is that someone is no longer with sales. It's the user with the marketing-view that should tell the database that this employee is now with marketing.

>> By the way, what is exactly your definition of "updatable"? Should there
>> be
>> at least one instance and one update such that the result is
>> well-defined, or should for all instances all updates be definied, or
>> something in between?

I assume you meant "Q * X = Y <=> X = Q^(-1) * Y" ?

That's a trivial property: every transformation is locally invertible in that sense. Note that because of the way you formulated it I only have to show that given an X an Y there is a Q^-1. An example would the function that maps every relation to X.

> #2 Transformation Q is globally invertible if #1 holds for any X and Y.

Since every transformation is always locally invertible for any X and Y it is also always globally invertable.

I doubt that this is what you meant and I'm not sure about what it is that you do mean.

> Next step is introducting deltas and demonstrating that locally invertible
> means updateable...

We cannot establish that if we don't have a definition of "updateable view", or do you accept mine? Anyway, we need a good definition of "invertible" first.

• Jan Hidders