Re: Notions of Type

From: Bob Badour <bbadour_at_pei.sympatico.ca>
Date: Thu, 17 Aug 2006 17:10:53 GMT
Message-ID: <xq1Fg.50732$pu3.588123@ursa-nb00s0.nbnet.nb.ca>

Marshall wrote:

> paul c wrote:
>

>>Marshall wrote:
>>
>>>erk wrote:
>>>
>>>>Sorry if this is obvious to everyone else, but does an algebra include
>>>>only operations defined on values of the type in question?
>>>
>>>Yes.
>>>
>>>
>>>>I ask
>>>>because in relational algebra, at least the rename operator involves a
>>>>different type ("attribute name") than the "core type" (relation).
>>>
>>>Very true. Of the various relational operators that have been
>>>identified over the years, only a few, like union, are really
>>>algebraic. RESTRICT, PROJECT, etc. aren't. ...
>>
>>Just a quick question - is your reason for making that statement that
>>restrict, project, etc. result in a relation that doesn't necessarily
>>have all the attributes of the operand relations?

>
>
> No; that's actually not a problem.
>
> The problem is the operands. Strictly speaking, an algebraic operator
> would be one which had the type
>
> op: 'a, 'a -> 'a
>
> ("The thing named "op" has the type: function taking operands of
> type some-a and some-a and returning a value of type some-a.")
>
> So the algebra of the integers has operators like:
>
> +: int, int -> int
>
> Consider PROJECT:
>
> PROJECT: Relation, Set-of-attributes -> Relation
>
> So for PROJECT of an x,y point over x, we pass it two
> things:
>
> 1) A relation defined over attributes x and y
> 2) ???
> and it returns
> 3) A relation defined over attribute x (aka "profit")
>
> 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. Exercise for the
> reader: what *is* the type of the other argument? This should
> make your head hurt a least a little bit.

It is a relation of degree 1 and cardinality N representing a set of N attribute names. Received on Thu Aug 17 2006 - 12:10:53 CDT