Normalization & Database Design: Finding Good Structure

1. 11. Normalization Database Design is the process of finding a "good" logical structure for the database. So far, design has been maintaining well-defined relations and choosing good primary keys. Here is a review of some of the definitions: A CANDIDATE KEY is a set of attributes, K, from the schema which always satisfies UNIQUENESS: no 2 distinct tuples have same K-value, MINIMALITY: None of Ai..Ak can be discarded and still maintain uniqueness. A PRIMARY KEY is one candidate key that is designated as the primary key (by a database administrator?). An ALTERNATE KEY is non-primary candidate key. A SUPERKEY is an attribute superset of a candidate key. NORMALIZATION is the part of Data Base design activities that precludes operational anomalies, and provide good key properties for projections. Normalization classes are (more exist): 1st Normal Form (1NF) 2nd Normal Form (2NF) 3rd Normal Form (3NF) BCNF Boyce Codd Normal Form (BCNF) 4th Normal Form (4NF) ...

2. The Database design expert? • NO! Not in isolation, anyway. • Someone from the enterprise who understands the data and the procedures should be consulted. • The following story illustrates this point. CAST: • Mr. Goodwrench = MG (parts manager); I ask again: Who decides primary key? (and other design choices?) • Pointy-headed Dbexpert = Ph D • Ph D Part Number will be the primary key for, PARTS(P#, COLOR, WT, TIME-OF-ARRIVAL). • MG You're the expert. • Ph D Well, P# should be the primary key, because IT IS the lookup attribute! . . . later • MG Why is lookup so slow? • Ph D You do store parts in the stock room ordered by P#, right? • MG No. We store by weight! When a shipment comes in, I take each part into the back room and throw it. The lighter ones go further than the heavy ones so they get ordered by weight! • Ph D But, but… weight doesn't have Uniqueness property! Parts with the same weight end up together in a pile! • MG No they don't. I tire quickly, so the first one I throw goes furthest. • Ph D Then we’ll use a composite primary key, (weight, time-of-arrival). • MG We get our keys primarily from Murt’s Lock and Key. Point: This conversation should have taken place during the 1st meeting.

3. Functional Dependencies The defining conditions get stronger going down this list and therefore the sets of qualifying relations gets smaller going down this list. By defining conditions: 4NF implies BCNF implies 3NF implies 2NF implies 1NF By classes of relations: 4NF is contained in BCNF is contained in 3NF is contained in 2NF is contained in 1NF Given a relation R with attributes X and Y (possibly composite) R.Y is FUNCTIONALLY DEPENDENT (FD) on R.X or equivalently, R.X FUNCTIONALLY DETERMINES R.Y written R.X → R.Y iff (x,y1) and (x,y2) in R[X,Y] implies y1=y2. NOTES: All attributes are FD on any attribute with the uniqueness property. Functional Dependencies are like integrity constraints. They are stipulated to hold (all tuples for all times) by designers. They can't be determined simply by observing the data state at a particular time. They are quite different from Association Rules. (ARs are approximate dependencies that hold at a given confidence level over a given subset at a particular time).

4. Full Functional Dependencies For composite attribute, X (composed of more than 1 attribute) R.Y is FULLY FUNCTIONALLY DEPENDENT (FFD) on R.X or R.X FULLY FUNCTIONALLY DETERMINES R.Y written R.XR.Y iff R.X→R.Y but R.Z→R.Y for any proper subset ZX. Said another way; R.X is a candidate key (has uniqueness and minimality) for R[X,Y]. These are SEMANTIC notions. One needs to know how data is used or what is intended (ask people who created/use data!)

5. First Normal Form (1NF) NORMAL FORMS (initially assuming only one candidate key) A file or relation is in First Normal Form (1NF) if there are no repeating groups (all attribute values are atomic). Allowing repeating groups in an attribute creates a situation in which the key does not determine a value in that attribute (but possibly many values). We can use a memory technique, that 1NF means: (note: THIS IS A MEMORY TECHNIQUE, NOT A DEFINITION - i.e., don't use this as a definition on an exam) every nonkey attribute is functionally dependent on key. This EDUCATION1 file is not in First Normal Form (1NF) S#|SNAME |CHILDREN|LCODE |LSTATUS|C#|CNAME|SITE |GR 32|THAISZ|Ed,Jo,Ty|NJ5102| 1 | 8| DSDE| ND |89 25|CLAY |Ann |NJ5101| 1 | 7| CUS | ND |68 32|THAISZ|Ed,Jo,Ty|NJ5102| 1 | 7| CUS | ND |91 25|CLAY |Ann |NJ5101| 1 | 6| 3UA | NJ |76 32|THAISZ|Ed,Jo,Ty|NJ5102| 1 | 6| 3UA | NJ |62 This EDUCATION2 file is in First Normal Form (1NF) S# |SNAME |LCODE |LSTATUS|C#|CNAME|SITE|GR 32|THAISZ|NJ5102| 1 | 8| DSDE| ND |89 25|CLAY |NJ5101| 1 | 7| CUS | ND |68 32|THAISZ|NJ5102| 1 | 7| CUS | ND |91 25|CLAY |NJ5101| 1 | 6| 3UA | NJ |76 32|THAISZ|NJ5102| 1 | 6| 3UA | NJ |62 How do you get EDUCATION1 into 1NF? 1. Create a separate CHILDREN file: CHILD|S# |Ed |32| |Jo |32| |Ty |32| |Ann |25|

6. Second Normal Form (2NF) A file or relation is in Second Normal Form (2NF) if it is in 1NF and every nonkey attribute is fully functionally dependent on the primary key.( memory scheme: every nonkey attribute is functionally dependent on whole key). Why do we need (want) relations to be in 2NF? 1NF relations which are not 2NF present PROBLEMS (anomalies), e.g., S# |SNAME |LCODE |LSTATUS|C#|CNAME|SITE|GR 32|THAISZ|NJ5102| 1 | 8| DSDE| ND |89 25|CLAY |NJ5101| 1 | 7| CUS | ND |68 32|THAISZ|NJ5102| 1 | 7| CUS | ND |91 25|CLAY |NJ5101| 1 | 6| 3UA | NJ |76 32|THAISZ|NJ5102| 1 | 6| 3UA | NJ |62 INSERT ANOMALY: Can't record JONES' LCODE until he takes a course. DELETE ANOMALY: Delete 1st record (e.g., THAISZ drops DSDE), loose C#=8 is DSDE in ND UPDATE ANOMALY: Change SITE of C#=6 from NJ to ND search sequentially for all C#=6.

7. Second Normal Form (cont.) E.g., In the EDUCATION1 file, with key (S#,C#), FD's which are not FFD are: (S#,C#) → SNAME (S#,C#) → LCODE (S#,C#) → LSTATUS (S#,C#) → CNAME (S#,C#) → SITE We make these FD's into FFD's by breaking (projecting) the file into 3 files. (puts relations in 2NF) STUDENTS = EDUCATION1[S#,SNAME,LCODE,LSTATUS] ENROLL = EDUCATION1[S#,C#,GRADE] COURSE = EDUCATION1[C#,CNAME,SITE] STUDENTS ENROLL COURSE |S#|SNAME |LCODE |LSTATUS| |S#|C#|GRADE| |C#|CNAME|SITE| |25|CLAY |NJ5101| 1 | |32|8 | 89 | |8 |DSDE |ND | |23|THAISZ|NJ5102| 1 | |32|7 | 91 | |7 |CUS |ND | |38|GOOD |FL6321| 4 | |25|7 | 68 | |6 |3UA |NJ | |17|BAID |NY2091| 3 | |25|6 | 76 | |5 |3UA |ND | |57|BROWN |NY2092| 3 | |32|6 | 62 | STUDENTS is in 2NF since it has a single attribute as key (S#) COURSE is 2NF since it has a single attribute as key (C#) ENROLL is 2NF since GRADE is FFD on (S#,C#) Still, we have problems: INSERT ANOMALY: Can't record LSTATUS of LOC=ND2987 until a student from ND2987 registers. DELETE ANOMALY: Deleting THAISZ, loose LCODE=NJ5102 has LSTATUS=1 UPDATE ANOMOLY: Change LSTATUS=3 to 2 and 4 to 3 must search STUDENTS sequentially. (so LSTATUS=2 isn't skipped!) The problem is: we have a transitive dependency: S#→LCODE LSTATUS, i.e., an FD which doesn't involve the key (or even part of the key): LCODE→LSTATUS A more common transitive dependency is "CityState": COURSE C# |CNAME|CTY|ST |8 |DSDE |Mot|ND| |7 |CUS |Mot|ND| |6 |3UA |Bay|NJ| |5 |3UA |Mot|ND| Delete 3rd record (cancel course C#=6) loose fact that Bay is in NJ, etc.

8. Third Normal Form (3NF) A file or relation is in Third Normal Form (3NF) if it is in 2NF and every nonkey attr is non-transitively dependent on primary key. (there are no transitive dependencies). (memory scheme: every nonkey attribute is functionally dependent on nothing but key) We need to project STUDENT ≡ STUDENTS[S#, SNAME, LCODE] and LOCATION ≡ STUDENTS[ LCODE, LSTATUS ] STUDENT LOCATION ENROLL COURSE S# |SNAME |LCODELCODE |LSTATUS|S#|C#|GRADE|C#|CNAME|SITE |25|CLAY |NJ5101 |NJ5101| 1 |32|8 | 89 |8 |DSDE |ND |23|THAISZ|NJ5102 |NJ5102| 1 |32|7 | 91 |7 |CUS |ND |38|GOOD |FL6321 |FL6321| 4 |25|7 | 68 |6 |3UA |NJ |17|BAID |NY2091 |NY2091| 3 |25|6 | 76 |5 |3UA |ND |57|BROWN |NY2092 |NY2092| 3 |32|6 | 62 In summary, the memory scheme to remember 3NF: (not a definition! Just a memory scheme) every nonkey attribute is functionally dependent upon the key 1NF the whole key and 2NF nothing but the key 3NF so help me Codd" (E.F. Codd invented relational model and normal forms) This is an analogy based on the way in which witnesses are sworn into legal proceedings in the US: Do you swear to tell the truth, the whole truth and nothing but the truth, so help you God?"

9. Boyce/Codd Normal Form (BCNF) A file or relation, with possibly multiple candidate keys, is in Boyce/Codd Normal Form (BCNF) if the only determinants are the superkeys (of candidate keys). Example of a 3NF relation which is not BCNF. ENROLL |S#|C#|GRADE|TUTOR| |32|8 | 89 |Xin | |32|7 | 91 |Sally| |25|7 | 68 |Ahmed| |25|6 | 76 |Ben | |32|6 | 62 |Amit | Primary key = (S#,C#) and each tutor is assigned to only 1 course. (Course to which a tutor is assigned is determined, so TUTOR → C# ) FDs: (S#,C#)→GRADE (S#,C#)→TUTOR (S#,C#)→(GRADE,TUTOR) TUTOR→C# This isn't BCNF (since Tutor is not superkey), but it is in 3NF (Strictly speaking, since C# is not a non-key attribute).

10. Fourth Normal Form (4NF) A file or relation, with possibly multiple candidate keys, is in Fourth Normal Form (4NF) if it is in BCNF and all Multivalue Dependencies (MVDs) are just FDs. What are Multivalue Dependencies? For R(A,X,Y), where A,X,Y are distinct attributes (possibly composite) R[A,X], R[A,Y] is a LOSSLESS decomposition of R iff R[A,X] JOINA R[A,Y] = R(A,X,Y) A set of projections of a relation with at a one common attribute and such that every attribute is in at least one projection is called a DECOMPOSITION. The join of a decomposition is always a superset of the original relation. Sometimes it is a proper superset (i.e., it includes SPURIOUS tuples that weren't in the original relation). R is always a subset of R[A,B] JOINA R[A,C] PROOF: If (a,b,c) is in R, (a,b) is in R[A,B] and (a,c) is in R[A,C]. Thus, (a,b,c) is in R[A,B] JOINA R[A,C]. Heath's theorem says when the join is exactly the original relation (a lossless decomposition). HEATH's THEOREM: Given R(A,B,C), if A→B or A→C then R = R[A,B] JOINA R[A,C] ie, if A→B or A→C then {R[A,B], R[A,C]} is a lossless decomposition Again: the join of a decomposition always contains the original relation but it may be larger (i.e., It may contain spurious tuples). Why call it a losswhen there are actually more tuples (extra spurious ones) and call it lossless when there is no gain in size? (next slide for answer).

11. Fourth Normal Form (4NF) cont. Proof of Heath's Theorem (Prove by contrapositive! i.e., P imples Q is true by showing NOT(Q) implies NOT(P) is true.) i.e., show NOT(R=R[A,B] JOINA R[A,C]) implies NOT(A→B or A→C ) = NOTA→B and NOTA→C (i.e., spurious tuples destroy at least one functional relationships) NOT(R = R[A,B] JOINA R[A,C]) means there exists an (a,b,c) in ( R[A,B] JOINA R[A,C] ) that is not in R (that is spurious) (a,b,c) in (R[A,B] JOINA R[A,C]) but (a,b,c) ∉ R implies (a,b) in R[A,B] and (a,c) in R[A,C] implies there are (a,b,c'), (a,b',c) in R such that c  c' and b  b' implies R.A does not functionally determine R.B and R.A does not functionally determine R.C QED. Does Heath's Theorem say: If R = R[A,B] JOINA R[A,C] then A→B (which would tell us that A→B implies A→C ) No! It is not an "if and only if". Counter example: R[A,B,C]: R[A,B]: R[A,C]: a b c a b a c a b' c a b' R[A,B] JOINA R[A,C] = R. But A NOT→ B. a b c a b' c So the FD A→B does not characterize lossless decomposition Is there a condition which does characterize lossless decomposition? i.e., an IF AND ONLY IF (IFF) condition for lossless decomposition (next slide)

12. Fourth Normal Form (4NF) cont. Yes, it is "Multivalued Dependency" or MVD: Given R(A,B,C), B is Multivalued Dependent on A (or A multi-determines B), written: A→→B, iff the set of B-values matching a given (a,c) pair in R depends only on the A-value, a (the same B-set is associated with a for all c's). Another way: A→→B iff forevery a in R.A, RR.A=a[B,C] (the projection onto B and C of the subscripted selection) is a product. If RR.A=a[B,C] is a product then clearly the set of B-values matching any pair, (a,c) is just the projection of that product onto the B-attribute and therefore A→→B. To prove A→→B implies RR.A=a[B,C] is a product, use the contrapositive arguement: If there is an A-value, a, such that RR.A=a[B,C] is not a product, R contains tuples: a b1 c1 a b1 c2 a b2 c1 (but R does not contain a b2 c2 ). But then, for c1, a→→b1,b2 but for c2, a→→b1 not the same set of B-values! MVD is a symmetric condition: Theorem: Given R(A,B,C), A→→B iff A→→C The proof follows directly from the previous result (the condition, RR.A=a[B,C] is a product and is symmetric in B and C).

13. Fourth Normal Form (4NF) cont. MVD is a symmetric condition: Theorem: Given R(A,B,C), A→→B iff A→→C The proof follows directly from the previous result (the condition, RR.A=a[B,C] is a product and is symmetric in B and C). Fagin's theorem: {R[A,B],R[A,C]} is a lossless decomposition of R(A,B,C) iff A→→B To prove A→→B implies decomposition is non-loss, prove contra-positive: decomposition is lossy implies A NOT→→ B If the decomposition is lossy, then there exists at least one (a,b) in R[A,B] and (a,c) in R[A,C] such that (a,b,c) ∉ R. Therefore, there is a b'  b in B such that (a,b',c) is in R and c'  c in C such that (a,b,c') is in R. Therefore pairs (a,c) and (a,c') do not determine the same B-sets in R (true since, b is in the B-set determined by (a,c') while it is not in the B-set determined by (a,c) ). To prove R=R[A,B] JOINA R[A,C] implies A→→B, prove the contra-positive: A NOT→→ B implies the decomposition is lossy. A NOT→→ B means there are distinct pairs (a,c) and (a,c') in R (and therefore in R[A,C]) which determine different B-sets in R, say b is in the B-set determined by (a,c) in R (and therefore (a,b,c) is in R) but b is not in the B-set determined by (a,c') in R (and therefore (a,b,c') is not in R), then (a,b,c') would not be in R. But since (a,b,c) is in R, (a,b) is in R[A,B] and (a,c') is in R[A,C] so (a,b,c') is in R[A,B] JOINA R[A,C] but not in R and the decomposition is lossy. 4th normal form (4NF)= BCNF and all MVDs are FDs (only dependencies are functional dependencies from superkeys).

14. The Normalization Process THE OVERALL NORMALIZATION PROCESS: 0. Project off repeating groups (each as separate files with repeating group attribute as key and the original key as foreign key) 1. Take projections of 1NF to eliminate nonfull FDs - produce 2NF. 2. Take projections of 2NF to eliminate transitive FDs - produce 3NF. 3. Take projections of 3NF to eliminate remaining FDs where determinant is not candidate key - produce BCNF. 4. Take projections of BCNF to eliminate any MVDs not FDs - produce 4NF. This discussion will stop with 4NF, however, there are 5NF, 6NF...18NF... Normalization has become an art form. CAUTION: It's not always completely clear what use will ever be made of some of these higher normal forms (that is; the anomolies being precluded are often somewhat involved and obscure).