How to find extraneous attribute? An example.


To find Extraneous attribute in a given functional dependency / How to find extraneous attribute? / Example for finding extraneous attributes

Finding Extraneous Attributes


Let us consider a set of functional dependencies F and any functional dependency of the form α → β. Assume that α and β are set of one or more attributes. [For example, A BC or AB CE etc.]
Case 1 (LHS): To find if an attribute A in α is extraneous or not. That is, to test if an attribute of Left Hand Side of a functional dependency is Extraneous or not.
Step 1: Find ({α} – A)+ using the dependencies of F.
Step 2: If ({α} – A)+ contains all the attributes of β, then A is extraneous.

Case 2 (RHS): To find if an attribute A in β is extraneous or not. That is, to test if an attribute of Right Hand Side of a functional dependency is Extraneous or not.
Step 1: Find α+ using the dependencies in F’ where F’ = (F – {α → β}) U { α → (β – A) }.
Step 2: If α+ contains A, then A is extraneous.

Example 1 for LHS:

Given F = {PQ, PQR}. Is Q extraneous in PQR?
In this example, we are looking for a LHS attribute. Hence, let us use Case 1 discussed above.
Step 1: Find ({α} – A)+ using the dependencies of F.
Here, α is PQ. So find (PQ – Q)+, ie., P+ (closure of P).
From F, if you know P, then you know Q (from PQ).
If you know both P and Q then you know R (from PQR).
Hence, the closure of P is PQR.
Step 2: If ({α} – A)+ contains all the attributes of β, then A is extraneous.
(PQ – Q)+ contains R. Hence, Q is extraneous in PQR.

Example 2 for RHS:

Given F = {PQR, QR}. Is R extraneous in PQR?
In this example, we are looking for a RHS attribute. Hence, let us use the Case 2 given above.
Step 1: Find α+ using the dependencies in F’ where F’ = (F – {α → β}) U { α → (β – A) }.
Let us find F’ as stated above.
F’ = (F – {α → β}) U { α → (β – A) } = ({PQR, QR} – {PQR}) U {P(QR-R)}
   = ({Q R} U {PQ})
F’ = {QR, PQ}
Here, α is P. So find (P)+, ie., closure of P using the F’ which we found.
From F, if you know P, then you know Q (from PQ).
If you know Q then you know R (from QR).
Hence, the closure of P is PQR.
Step 2: If α+ contains A, then A is extraneous.
P+ contains R. Hence, R is extraneous in PQR.



2 comments:

  1. Did you mean to say F' in example 2 step 1 when you said "from F, if you know P, then you know Q"?

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    Replies
    1. Thanks for your comment. Now I have changed to F' in example 2.

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