Re: atomic
From: paul c <toledobythesea_at_ooyah.ac>
Date: Tue, 06 Nov 2007 03:02:49 GMT
Message-ID: <tPQXi.178630$Da.71130_at_pd7urf1no>
>
> r1:
>
> Name Car
> ----------- ------------
> {bill} {car1,car2,car4}
> {john,fred} {car3}
>
>
> r2:
>
> Car Colour
> ---------------- ---------
> {car1,car3,car4} {red}
> {car2} {green}
>
>
> r1 join ( r2 rename Car as Vehicle ):
>
> Name Car Vehicle Colour
> ------------- ------------------ ----------------- ---------------
> {bill} {car1,car2,car4) {car1,car3,car4} {red}
> {bill} {car1,car2,car4) {car2} {green}
> {john,fred} {car3} {car1,car3,car4} {red}
> {john,fred} {car3} {car2} {green}
>
> I count 4 rows.
Date: Tue, 06 Nov 2007 03:02:49 GMT
Message-ID: <tPQXi.178630$Da.71130_at_pd7urf1no>
Bob Badour wrote:
> paul c wrote:
>
>> Bob Badour wrote: >> >>> paul c wrote: >>> >>>> paul c wrote: >>>> ... >>>> >>>>> if those make sense, by the D&D definition of "cartesian product", >>>>> I think r1 |x| r2 would be empty and I presume so would I(r1 |x| r2). >>>> >>>> Sorry, let me take that back, I think I should have said >>>> intersection, not cartesian product. >>> >>> The join is empty. The intersection is empty. If one renamed one of >>> the "car" attributes to something else, the join would have 4 rows. >> >> I think it would have 20!
>
> r1:
>
> Name Car
> ----------- ------------
> {bill} {car1,car2,car4}
> {john,fred} {car3}
>
>
> r2:
>
> Car Colour
> ---------------- ---------
> {car1,car3,car4} {red}
> {car2} {green}
>
>
> r1 join ( r2 rename Car as Vehicle ):
>
> Name Car Vehicle Colour
> ------------- ------------------ ----------------- ---------------
> {bill} {car1,car2,car4) {car1,car3,car4} {red}
> {bill} {car1,car2,car4) {car2} {green}
> {john,fred} {car3} {car1,car3,car4} {red}
> {john,fred} {car3} {car2} {green}
>
> I count 4 rows.
Okay, I thought you were using the sources from David B's post about his "I" function which would have 20. Received on Tue Nov 06 2007 - 04:02:49 CET