1. cs3431 Normalization

2. cs3431 Why Normalization? To remove potential redundancy in design Redundancy causes several anomalies: insert, delete and update Redundancy wastes storage, and often slows down query processing Examples to follow next.

3. cs3431 What is Normalization? Normalization uses concept of dependencies Functional Dependencies Technique used: Decomposition Break R (A, B, C, D) into R1 (A, B) and R2 (B, C, D)

4. cs3431 Insert Anomaly sNumbersNamepNumberpName s1Davep1MM s2Gregp2ER Student Question: Could we insert any professor ? Note: We cannot insert a professor who has no students. Insert Anomaly: We are not able to insert “valid” value/(s)

5. cs3431 Delete Anomaly sNumbersNamepNumberpName s1Davep1MM s2Gregp2ER Student Question: Can we delete a student and keep a professor info ? Note: We cannot delete a student that is the only student of a professor. Note: In both cases, minimum cardinality of Professor in the corresponding ER schema is 0 Delete Anomaly: We are not able to perform a delete without losing some “valid” information.

6. cs3431 Update Anomaly sNumbersNamepNumberpName s1Davep1MM s2Gregp1MM Student Question: Can we simply update a professor’s name ? Note: To update the name of a professor, we have to update in multiple tuples. Note the maximum cardinality of Professor in the corresponding ER schema is * Update Anomaly: To update a value, we have to update multiple rows. Update anomalies are due to redundancy.

7. cs3431 Normalization Need a method to find such “dependencies” exist between attributes Functional dependencies Need a method to remove such harmful dependencies, when they exist Relational decomposition

8. cs3431 Keys : Revisited A key for a relation R (a1, a2, …, an) is a set of attributes, K, that together uniquely determine the values for all attributes of R. A key is minimal: no subset of K is a key. A superkey need not be minimal A prime attribute: an attribute that is part of a key

9. cs3431 Keys: Example sNumbersNameaddress 1Dave144FL 2Greg320FL Student Primary Key: Candidate key: Some superkeys: {,, } Prime Attribute: {sNumber, sName}

10. cs3431 Functional Dependencies (FDs) sNumbersNameaddress 1Dave144FL 2Greg320FL Student Suppose we have the FD: sName  address That is, there is a function from sName to address Meaning: For any two rows in the Student relation with the same value for sName, the value for address must be same.

11. cs3431 FD and Keys sNumbersNameaddress 1Dave144FL 2Greg320FL Student Questions : Does a key implies functional dependencies? Which ones ? Does a functional dependency imply keys ? Which ones ? Observation : Any key (primary or candidate) or superkey of a relation R functionally determines all attributes of R. Primary Key : FD : sName  address

12. cs3431 Properties of FDs Consider A, B, C, Z are sets of attributes Reflexive (trivial FD): if A  B, then A  B Transitive: if A  B, and B  C, then A  C Augmentation: if A  B, then AZ  BZ Union: if A  B, A  C, then A  BC Decomposition: if A  BC, then A  B, A  C Note: Sound and complete inference rules for FDs

13. cs3431 Inferring FDs Suppose we have : a relation R (A, B, C) and functional dependencies A  B, B  C, C  A Questions : What is a key for R? Should we split R into multiple relations? We can infer A  ABC, B  ABC, C  ABC. Hence A, B, C are all keys.

14. cs3431 Reasoning About FDs An FD f is implied by a set of FDs F if f holds whenever all FDs in F hold. = closure of F is the set of all FDs that are implied by F. Computing closure of a set of FDs can be expensive. Size of closure is exponential in # attrs!

15. cs3431 Reasoning About FDs Instead of computing closure F+ of a set of FDs Too expensive Typically, we just need to know if a given FD X Y is in closure of a set of FDs F. Algorithm for efficient check: Compute attribute closure of X (denoted ) wrt F: Set of all attributes A such that X A is in There is a linear time algorithm to compute this. Check if Y is in X+. If yes, then X  A in F+.

16. cs3431 Reasoning About FDs (Contd.) Does F = {A B, B C, C D E } imply A E? Question : i.e, is A E in the functional set closure ? Equivalent Question : Is E in the attribute closure ?

17. cs3431 Algorithm for Inference of FDs Computing the closure of set of attributes {A1, A2, …, An}, denoted {A1, A2, …, An} + 1. Let X = {A1, A2, …, An} 2. If there exists a FD : B1, B2, …, Bm  C, such that every Bi  X, then X = X  C 3. Repeat step 2 until no more attributes can be added. 4. {A1, A2, …, An} + = X

18. cs3431 Inferring FDs: Example 1 Consider R (A, B, C, D, E) with FDs A  B, B  C, CD  E Does A  E? (Is A  E in F+ ?) Rephrase as : Is E in A+ ? Let us compute {A} + {A} + = {A, B, C} Therefore, A  E is false

19. cs3431 Inferring FDs: Example 2 Given R (A, B, C), and FDs : A  B, B  C, C  A What are possible keys for R ? Compute the closure of attributes: {A} + = {A, B, C} {B} + = {A, B, C} {C} + = {A, B, C} So keys for R are,,

20. cs3431 Decomposing Relations sNumbersNamepNumberpName s1Davep1MM s2Gregp2MM StudentProf FDs: pNumber  pName sNumbersNamepNumber s1Davep1 s2Gregp2 Student pNumberpName p1MM p2MM Professor

21. cs3431 Decomposition Decomposition: Must be Lossless (no spurious tuples)

22. cs3431 Decomposition: Lossless Join Property sNumbersNamepName S1DaveMM S2GregMM Student pNumberpName p1MM p2MM Professor sNumbersNamepNumberpName s1Davep1MM s1Davep2MM s2Gregp1MM s2Gregp2MM StudentProf Spurious Tuples

23. cs3431 Normalization How do I decide if I need to further decompose? Once decided, what is the algorithm for decomposing?