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Canonical Cover


Whenever a user updates the database, the system must check whether any of the functional dependencies are getting violated in this process. If there is a violation of dependencies in the new database state, the system must roll back. Working with a huge set of functional dependencies can cause unnecessary added computational time. This is where the canonical cover comes into play.
A canonical cover of a set of functional dependencies F is a simplified set of functional dependencies that has the same closure as the original set F.

Important definitions:

Extraneous attributes: An attribute of a functional dependency is said to be extraneous if we can remove it without changing the closure of the set of functional dependencies.
Canonical cover: A canonical cover F_{c} of a set of functional dependencies F such that ALL the following properties are satisfied:

  • F logically implies all dependencies in F_{c}.
  • F_{c} logically implies all dependencies in F.
  • No functional dependency in F_{c} contains an extraneous attribute.
  • Each left side of a functional dependency in F_{c} is unique. That is, there are no two dependencies alpha_{1} 
ightarrow eta_{1} and alpha_{2} 
ightarrow eta_{2} in such that alpha_{1} 
ightarrow alpha_{2}.

Finding Canonical Cover

Algorithm to compute canonical cover of set F:



repeat
    1. Use the union rule to replace any dependencies in 
       alpha_{1} 
ightarrow eta_{1} and alpha_{1} 
ightarrow eta_{2} with alpha_{1} 
ightarrow eta_{1}eta_{2}.
    2. Find a functional dependency alpha 
ightarrow eta with an 
       extraneous attribute either in alpha or in eta.
    3. If an extraneous attribute is found, delete it from alpha 
ightarrow eta.
       until F does not change

Example1:

Consider the following set F of functional dependencies:

F= {
A 
ightarrow BC
B 
ightarrow C
A 
ightarrow B
AB 
ightarrow C
}

Steps to find canonical cover:

  1. There are two functional dependencies with the same set of attributes on the left:
    A 
ightarrow BC
    A 
ightarrow B

    These two can be combined to get
    A 
ightarrow BC.

    Now, the revised set F becomes:

    F= {
    A 
ightarrow BC
    B 
ightarrow C
    AB 
ightarrow C
    }

  2. There is an extraneous attribute in AB 
ightarrow C because even after removing AB 
ightarrow C from the set F, we get the same closures. This is because B 
ightarrow C is already a part of F.

    Now, the revised set F becomes:

    F= {
    A 
ightarrow BC
    B 
ightarrow C
    }



  3. C is an extraneous attribute in A 
ightarrow BC, also A 
ightarrow B is logically implied by A 
ightarrow B and B 
ightarrow C (by transitivity).

    F= {
    A 
ightarrow B
    B 
ightarrow C
    }

  4. After this step, F does not change anymore. So,
    Hence the required canonical cover is,
    F_{c}= {
    A 
ightarrow B
    B 
ightarrow C
    }

Example2:

Consider another set F of functional dependencies:

F={
A 
ightarrow BC
CD 
ightarrow E
B 
ightarrow D
E 
ightarrow A
}

  1. The left side of each functional dependency in F is unique.
  2. None of the attributes in the left or right side of any functional dependency is extraneous (Checked by applying definition of extraneous attributes on every functional dependency).
  3. Hence, the canonical cover F_{c} is equal to F.

Note: There can be more than one canonical cover F_{c} of a set F of functional dependencies.

How to check whether a set of f.d.’s F canonically cover another set of f.d.’s G?

Consider the following two sets of functional dependencies:

F = {
A 
ightarrow B
AB 
ightarrow C
D 
ightarrow AC
D 
ightarrow E
}

G = {
A 
ightarrow BC
D 
ightarrow AB
}



Now, we are required to find out whether one of these f.d.’s canonically covers the other set of f.d.’s. This means, we need to find out whether F canonically covers G, G canonically covers F, or none of the two canonically cover the other.

To find out, we follow the following steps:

  • Create a singleton right hand side. This means, the attributes to the right side of the f.d. arrow should all be singleton.
    The functional dependency D 
ightarrow AC gets broken down into two functional dependencies, D 
ightarrow A and D 
ightarrow C.

    F = {
    A 
ightarrow B
    AB 
ightarrow C
    D 
ightarrow A
    D 
ightarrow C
    D 
ightarrow E
    }

  • Remove all extraneous attributes.

    Consider any functional dependency XY 
ightarrow Z. If X in itself can determine Z, then the attribute Y is extraneous and can be removed. As we can see, the occurrence of extraneous attributes is possible only in those functional dependencies where there are more than one attributes in the LHS.

    So, consider the functional dependency AB 
ightarrow C.
    Now, we must find the closures of A and B to find whether any of these is extraneous.

    [A]^{+}=AB
    [B]^{+}=B

    As we can see, B can be determined from A. This means we can remove B from the functional dependency AB 
ightarrow C.

    F = {
    A 
ightarrow B
    A 
ightarrow C
    D 
ightarrow A
    D 
ightarrow C
    D 
ightarrow E
    }

  • Remove all redundant functional dependencies.

    Check all f.d.’s one by one, and see if by removing a f.d. X 
ightarrow Y, we can still find out Y from X by some other f.d. A more formal way to state this is find [X]^{+} without making use of the f.d. we are testing and check whether Y is a part of the closure. If yes, then the f.d. is redundant.



    Here, when checking for the f.d. D 
ightarrow C, we observe that even after hiding it, the closure of D contains C. This is because we can obtain C from D by the combination of two other f.d.’s D 
ightarrow A and A 
ightarrow C. So, 
ightarrow C is redundant.

    F = {
    A 
ightarrow B
    A 
ightarrow C
    D 
ightarrow A
    D 
ightarrow E
    }

Now, do the same for G.

  • Create a singleton right hand side. This means, the attributes to the right side of the f.d. arrow should all be singleton.

    G = {
    A 
ightarrow B
    A 
ightarrow C
    D 
ightarrow A
    D 
ightarrow B
    }

  • Remove all extraneous attributes.
    Since the RHS of all f.d.’s contains only 1 attribute, there is no extraneous attribute possible.

  • Remove all redundant functional dependencies.

    By looping over all f.d.’s and checking the closure of the LHS in all cases, we observe that the f.d. D 
ightarrow B is redundant as it can be obtained through a combination of 2 other f.d.’s, D 
ightarrow A and A 
ightarrow B.

    G = {
    A 
ightarrow B
    A 
ightarrow C
    D 
ightarrow A
    }

Now, since all f.d.’s of G are already covered in F, we conclude that F covers G.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.



This article is attributed to GeeksforGeeks.org

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