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: Attention reader! 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 of a set of functional dependencies F such that ALL the following properties are satisfied:. Finding Canonical Cover Algorithm to compute canonical cover of set F: repeat 1. Use the union rule to replace any dependencies in and with. Find a functional dependency with an extraneous attribute either in or in.
If an extraneous attribute is found, delete it from. 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. Hence, the canonical cover is equal to F. Note: There can be more than one canonical cover of a set F of functional dependencies. How to check whether a set of f. 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. Consider any functional dependency XY 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 C. Improve this question. Mitch Mitch 71 1 1 gold badge 1 1 silver badge 2 2 bronze badges. Add a comment. Active Oldest Votes.
Improve this answer. Sign up or log in Sign up using Google. Sign up using Facebook. Sign up using Email and Password. Post as a guest Name. Email Required, but never shown. The Overflow Blog. Podcast The first ten years of our programming lives. Upcoming Events. No relation contains both attributes. Step 2. If no relation in D contains a candidate key of R, create a new relation with a candidate key of R being the schema and add it to the result D.
There is only one candidate key of R: AD. Simplify D by removing relations that are redundant i. R is not in 2NF. It is not possible to decompose R into two relations losslessly while preserving all functional dependencies. However, it is possible to decompose into three relations losslessly and with all functional dependencies preserved:. If the department an employee is working on is independent of the skill that he has, there is redundancy. For example, the fact that employee has the skill PHP is stored twice.
Suppose we have the following relation instance under a different set of assumptions:. Consider Example 1 again. In the data modeling, there may be two classes with a many-to-many association. If one follows the data modeling and the mapping to relation guideline, two relations will be created instead of one. It should be decomposed into:. We have:. An employee provides some skills that a project needs. If an employee has a skill that is not needed by the project e.
It does not show up in R because it is embedded. We may have the following three classes with three many-to-many associations between each pair of them:. Decomposition of R into R1 and R2 will lose this embedded multivalued dependency. There are many to many binary associations between A and C as well as B and C.
Furthermore, there is a ternary association between A, B and D. Both relations are now in DKNF.
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