Re: Notions of Type

Marshall wrote:
> Whoops! Doesn't fit the template. The second argument isn't
> a relation. So, strictly speaking, this is not an algebraic operator,
> because it isn't closed over the type Relation.

To be pedantic, do operators of type R x R x ... -> R x R x ..., for any number of R on either side, fit the algebra of R? Or is there some cardinality "constraint"?

> Exercise for the
> reader: what *is* the type of the other argument? This should
> make your head hurt a least a little bit.

Maybe I'm off-base, but isn't the type of the second argument a subset of the heading of the relation? So if RELATION is a type generator, the second argument is a type parameterized by the actual type of the relation parameter. There are various sense of type parameterization here, and I'm not smart enough right now to find their proper names. Does some type jockey care to rescue us?

> Note that the type Relation is a what Date et al call a "type
> generator"
> and others call a parameterized type.

Are these one and the same?

> [...] I think we've all intuited, loosely
> speaking, that there's a real algebra in there trying to get out.

I think for purposes of the relational model we all know and love (especially its useful operators), a real algebra won't suffice. Or am I just being short-sighted?

• erk