Database Design= finding a "good" logical structure for the database. So far, design criteria has been just maintaining well-defined relations and choosing good primary keys. Review of some definitions:CANDIDATE KEYis a set of attributes, K from the schema which, at all times, satisfies: UNIQUENESS: no 2 distinct tuples have same K-value, MINIMALITY: None of Ai..Ak can be discarded and still maintain uniqueness.PRIMARY KEY= one candidate key designated primaryALTERNATE KEY= non-primary candidate keySUPERKEY= attribute superset of a candidate keyNORMALIZATION= DB design activities to preclude 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) ... 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 ⇒ BCNF ⇒ 3NF ⇒ 2NF ⇒ 1NF By classes of relations 4NF ⊆ BCNF ⊆ 3NF ⊆ 2NF ⊆ 1NF Given a relation R with attributes X and Y (possibly composite) R.Y isFUNCTIONALLY DEPENDENT (FD)on R.X equivalently, R.XFUNCTIONALLY DETERMINESR.Y writtenR.X → R.Yiff (x,y_{1}) and (x,y_{2}) ∈ R[X,Y] implies y_{1}=y_{2}. 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 For composite attribute, X (composed of more than 1 attribute) R.Y isFULLY FUNCTIONALLY DEPENDENT (FFD)on R.X R.XFULLY FUNCTIONALLY DETERMINESR.Y written R.X ⇒ R.Y iff R.X → R.Y but R.Z not→ 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!)(initially assuming only one candidate key)NORMAL FORMS= 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 it on the comprehensive!)First Normal Form (1NF)This EDUCATION1 file is not in First Normal Form (1NF)every nonkey attribute is functionally dependent on keyS# SNAME CHILDREN LCODE LSTATUS C# CNAME SITE GR32|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 GR32|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| or 2. Assuming a maximum number of children, say 3:S# SNAME CHILD1 CHILD2 CHILD3 LCODE LSTATUS C# CNAME SITE GR32|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= 1NF and every nonkey attribute is fully functionally dependent on the primary key.Second Normal Form (2NF)Why do we need (want) relations to be in 2NF? 1NF relations which are not 2NF present PROBLEMS (anomalies):every nonkey attribute is functionally dependent on whole keyS# SNAME LCODE LSTATUS C# CNAME SITE GR32|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# from NJ to ND search sequentially for all C#=6 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[S#,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 LSTATUS=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., FD which doesn't involve key (or part of key): LCODE → LSTATUS A more common transitive dependency is "City-State":COURSEC# CNAME CTY ST|8 |DSDE |Mot|ND| |7 |CUS |Mot|ND| |6 |3UA |Bay|NJ| |5 |3UA |Mot|ND| Delete 3^{rd}record (cancel course C#=6) loose fact that Bay is in NJ, etc.: 2NF and every nonkey attr is non-transitively dependent on primary key. (there are no transitive dependencies).Third Normal Form (3NF)We need to project STUDENT ≡ STUDENTS[S#,SNAME,LCODE] and LOCATION ≡ STUDENTS[LCODE,LSTATUS]every nonkey attribute is functionally dependent on nothing but keySTUDENTLOCATIONENROLLCOURSES# SNAME LCODE LCODE 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)This is an analogy based on the way in which witnesses are sworn into legal proceedings in the US: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)Do you swear to tell the truth, the whole truth and nothing but the truth, so help you God?"DETERMINANT = any attribute on which some other attribute is functionally dependent. The "The key"-part implies there is only one candidate key. A more general normal form is more explicit with respect to alternate keys:: The only determinants are superkeys (superset of candidate key) Example of a 3NF relation which is not BCNF.BCNF (Boyce/Codd Normal Form)ENROLLS# 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 attr.: BCNF and all Multivalue Dependencies (MVDs) are 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 a4^{th}normal form (4NF)LOSSLESS decompositionof R iffR[A,X]A-JOINR[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 aDECOMPOSITIONThe join of a decomposition is always a superset of the original relation. Sometimes it is a proper superset (i.e., it includestuples that weren't in the original relation). R is always a subset ofSPURIOUSR[A,B]A-JOINR[A,C]PROOF: ∀ a,b,c ∈ R (a,b)∈R[A,B] and a,c∈R[A,C]. Thus, a,b,c∈ R[A,B] A-JOIN R[A,C]. Heath's theorem says when the join is exactly the original relation (a lossless decomposition).: Given R(A,B,C), if A→B or A→C then R = R[A,B] A-JOIN 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 (contain spurious tuples) Why call it aHEATH's THEOREMwhen there are actually more tuples (extra spurious ones) and call itlosswhen there is no gain in size?losslessProof of Heath's Theoremby contrapositive (Provetrue by showingP imples Qtrue) ie, show NOT(R=R[A,B] A-JOIN 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] A-JOIN R[A,C]) means ∃ (a,b,c) ∈ ( R[A,B] A-JOIN R[A,C] ) not in R (spurious) (a,b,c) ∈ (R[A,B] JOIN R[A,C]) but ∉ R implies (a,b) ∈ R[A,B] and (a,c) ∈ R[A,C] implies ∃ (a,b,c'), (a,b',c) ∈ R∋: c NOT= c' and b NOT= 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] A-JOIN 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: R[A,B]: R[A,C]: a b c a b a c a b' c a b' R[A,B] JOIN R[A,C] = R But A NOT→ B. a b c a b' c So the FD A→B does notNOT(Q) implies NOT(P)lossless decomposition Is there a condition which does characterize lossless decomposition? (i.e., an IF AND ONLY IF (IFF) condition for lossless decomposition) Yes, it is "Multivalued Dependency" or MVD: Given R(A,B,C),characterize(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 ∀ a∈R.A, RB is Multivalued Dependent on A_{R.A=a}[B,C] is a product. If R_{R.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 R_{R.A=a}[B,C] is a product, use the contrapositive arguement: If there is an A-value, a, ∋: R_{R.A=a}[B,C] is not a product then 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 Lemma (the condition, R_{R.A=a}[B,C] is a product is symmetric in B and C).Fagin's thm: {R[A,B],R[A,C]} is a lossless decompostion of R(A,B,C) iff A→→B Proof: To prove A→→B implies the decomposition is nonloss, prove the contrapositive: the decomposition is lossy implies A NOT→→ B. If the decomp is lossy, then there exists at least one (a,b) ∈ R[A,B] and (a,c) ∈ R[A,C] ∋: (a,b,c) ∉ R. Therefore, there is b' not= b in B such that (a,b',c) is in R and a c' not= 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 (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]joinR[A,C] implies A→→B, prove the contrapositive: A NOT→→ B implies the decomp is lossy. A NOT→→ B means there are distinct pairs a,c and a,c' ∈ 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' is not 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] JOIN R[A,C] but not in R and the decomposition is lossy.= BCNF and all MVDs are FDs (only dependencies are functional dependencies from superkeys) 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. 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). some powerpoint notes on normalisation4th normal form (4NF)