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Re: circular relationships ok?

From: Jan Hidders <jan.hidders_at_REMOVETHIS.pandora.be>
Date: Sun, 05 Mar 2006 12:18:20 GMT
Message-ID: <gGAOf.296091$IR1.9311296@phobos.telenet-ops.be>


Brian Selzer wrote:
>
> In other words, a database schema consisting of the relation schemata,
>
> R{A, B}, S{B, C}, and T{C, A}
>
> and the foreign key constraints,
>
> R(B) --> S(B), S(C) --> T(C), T(A) --> R(A)
>
> is equivalent to a database schema consisting of the same relation schemata
> and the foreign key constraints,
>
> R(A) --> T(A), T(C) --> S(C), S(B) --> R(B)
>
> is equivalent to a database schema consisting of the relation schema
>
> U(A, B, C) where A, B, and C are candidate keys.

Actually, none of them are equivalent. Consider the following database instances: (formatted assuming fixed width font)

R : A B S : B C T : C A

This is an instance of the first schema, but not of the second. A similar example shows thre is an instance of the second schema that is not an instance of the first schema:

R : A B S : B C T : C A

Finally, it is clear that these two instances cannot be represented in the third schema since the join would lose tuples. So they are really all three different.

What *is* true is that the following schema:

Is equivalent with the schema:

So perhaps that is what you meant?

Received on Sun Mar 05 2006 - 06:18:20 CST

Original text of this message

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