Flashcards
Normalization Theory
Cards in this deck (15)
- 1What are some of the limitations of relational database design?- no unique database schema
- repetition of information
- loss of information - 2What causes redundancy?Dependencies between attributes
e.g. all addresses in the same town have the same postal code - 3What is an update anomaly?When a change in some field must be made in several places as well
- 4What is a functional dependency?A functional dependency on a relation schema R is a constraint X --> Y, where X and Y are subsets of attributes of R.
- 5State Armstrong's Reflexivity axiom.If Y ⊆ X then X --> Y (trivial FD)
- 6State Armstrong's Augmentation axiom.If X --> Y then XZ --> YZ
- 7State Armstrong's Transitivity axiom.If X --> Y and Y --> Z, then X --> Z.
- 8What is an attribute closure?If X is a set of attributes, its attribute closure is the set of all attributes, A, such that X --> A.
- 9Consider the functional dependency F: AB --> C, A --> D, D --> E, AC --> B Is AB --> E entailed by F? Is D --> C entailed by F?Yes
No - 10Compute the attribute closure of AB with respect to the set of FDs: AB --> C, A --> D, D --> E, AC --> BThe attribute closure of AB is {ABCDE}.
- 11When is a relation schema R said to be in BCNF (Boyce-Codd Normal Form)?A relation schema R is in BCNF if for every FD X --> Y associated with R, either
- Y ⊆ X (i.e. the FD is trivial), or
- X is a superkey of R - 12Consider the schema Person (SIN, Name, Address, Hobby) with key (SIN, Hobby). Does the FD SIN --> Name, Address satisfy the requirements of BCNF?No, it does not
- 13True or False: BCNF does not eliminate redundancy.False
- 14True or False: Decomposition must be lossless.True
- 15Decompose the following relation with FDs into BCNF: CSJDPQV, key C, JP -> C, SD -> P, J -> STo deal with SD -> P, decompose into SDP, CSJDQV.
To deal with J -> S, decompose CSJDQV into JS and CJDQV.
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- File and Database Management Flashcards on Normalization Theory, created by Chuck H on 13/04/2019.
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- Chuck H
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