Re: constraints in algebra instead of calculus
Date: Fri, 15 Jun 2007 13:42:40 -0300
Message-ID: <4672c17b$0$4312$9a566e8b_at_news.aliant.net>
Jan Hidders wrote:
> On 15 jun, 16:32, Bob Badour <bbad..._at_pei.sympatico.ca> wrote:
>
>>Jan Hidders wrote:
>>
>>>On 15 jun, 04:46, paul c <toledobythe..._at_oohay.ac> wrote:
>>
>>>>paul c wrote:
>>
>>>>>paul c wrote:
>>
>>>>>>...
>>
>>>>>>3) Use TTM-style GROUP/UNGROUP to test that
>>>>>>UNGROUP {NK}
>>>>>> (GROUP ( (A GROUP {NK} as NK ))
>>>>>>= A
>>>>>>...
>>
>>>>>Oops, that last one was completely nuts, can't remember now whatever I
>>>>>was thinking. Sorry for wasting people's time.
>>
>>>>>p
>>
>>>>Marshall, this mis-step has bugged me for days, for no reason that I can
>>>>explain, except that at one time it all seemed crystal-clear and I'm not
>>>>sure now whether my memory is bad or whether I was flat-out wrong to
>>>>think of GROUP et cetera in this way. This might be what I was trying
>>>>to remember:
>>
>>>>(Assuming D&D's Tutorial D syntax which I've probably mangled but hoping
>>>>you get the drift. I've tried to write it slow-motion with a bunch of
>>>>relvars and assignments, because I don't think I could avoid those
>>>>without error, even when wide-awake and sober.)
>>
>>>>R is relvar name, A is candidate key, B is set of all the non-key
>>>>attributes.
>>
>>>>1. R1 := R GROUP ( {A} as gA )
>>>>2. R2 := R1 GROUP ( {B} as gB)
>>>>3. R3 := R GROUP ( {B} as gB )
>>>>4. R4 := R3 GROUP ( {A} as gA )
>>>>5. R5 := R4{gB} <AND> R2
>>>>6. R6 := UNGROUP ( R5 UNGROUP (gB)) (gA) )
>>>>7. (constraint) R6 = R
>>
>>>You can simplify that a bit:
>>
>>>R1 := (R{B}) GROUP {B} AS gB
>>>R2 := (R GROUP {B} AS gB){gB}
>>>constraint: R1=R2
>>
>>That constraint looks like a tautology to me. Can you explain how any
>>relation with a B attribute could fail the constraint?
>
>
> I'm not that well versed in the TTM / Tutorial D notation so I may
> have abused the notation a bit. To clarify:
> - B is a set of attributes (the non-key attributes)
> - R{B} denotes the projection of R on the attributes in B
> - R GROUP {B} AS C groups the attributes in B and names the resulting
> set-valued attribute C
>
> That probably clears it up, but I'll give a small example anyway
> (note: here B is not a set of attributes but a single attribute):
>
> Assume R = { (A:1, B:2), (A:1, B:3) }
>
> R{B} = { (B:2}, (B:3) }
> R1 = R{B} GROUP {B} AS gB = { (gB:{ (B:2} }), (gB:{ (B:3) }) }
>
> (R GROUP {B} AS gB) = { (A:1, gB:{ (B:2}, (B:3) }) }
> R2 = (R GROUP {B} AS gB){gB} = { (gB:{ (B:2}, (B:3) }) }
>
> So you see that in this case R1 and R2 are not equal, which indicates
> that {A} is not a superkey in R.
>
> -- Jan Hidders
Thanks Jan! Now I see it! Received on Fri Jun 15 2007 - 18:42:40 CEST