Revision for Midterm 3 revision 3 Prof. Sin-Min Lee Department of Computer Science Set Operators • Relation is a set of tuples, so set operations should apply: , , (set difference) • Result of combining two relations with a set operator is a relation; hence all its elements must be tuples having the same structure • Hence, scope of set operations limited to union compatible relations Union Compatible Relations • Two relations are union compatible if – Both have same number of columns – Names of attributes are the same in both – Attributes with the same name in both relations have the same domain • Union compatible relations can be combined using union, intersection, and set difference Example Tables: Person (SSN, Name, Address, Hobby) Professor (Id, Name, Office, Phone) are not union compatible. But Name (Person) and Name (Professor) are union compatible so Name (Person) makes sense. - Name (Professor) Cartesian Product • If R and S are two relations, R S is the set of all concatenated tuples <x,y>, where x is a tuple in R and y is a tuple in S (but see naming problem next) • R S is expensive to compute: – Factor of two in the size of each row – Quadratic in the number of rows A B x1 x2 x3 x4 C D y1 y2 y3 y4 R S A x1 x1 x3 x3 B C x2 y1 x2 y3 x4 y1 x4 y3 R S D y2 y4 y2 y4 Renaming in Cartesian Product Result of expression evaluation is a relation. Attributes of relation must have distinct names. So what do we do if they don’t? E.g., suppose R(A,B) and S(A,C) and we wish to compute R S . One solution is to rename the attributes of the answer: R S( R.A, R.B, S.A, S.C) Although only A needs to be renamed, it is“cleaner” to rename them all. Renaming Operator • Previous solution is used whenever possible but it won’t work when R is the same as S. • Renaming operator resolves this. It allows to assign any desired names, say A1, A2,… An , to the attributes of the n column relation produced by expression expr with the syntax expr [A1, A2, … An] Example Transcript (StudId, CrsCode, Semester, Grade) Teaching (ProfId, CrsCode, Semester) StudId, CrsCode (Transcript)[StudId, CrsCode1] ProfId, CrsCode(Teaching) [ProfId, CrsCode2] This is a relation with 4 attributes: StudId, CrsCode1, ProfId, CrsCode2 Derived Operation: Join A (general or theta) join of R and S is the expression R join-condition S where join-condition is a conjunction of terms: Ai oper Bi in which Ai is an attribute of R; Bi is an attribute of S; and oper is one of =, <, >, , . The meaning is: join-condition´ (R S) where join-condition and join-condition´ are the same, except for possible renamings of attributes caused by the Cartesian product. Theta Join – Example Employee(Name,Id,MngrId,Salary) Manager(Name,Id,Salary) Output the names of all employees that earn more than their managers. Employee.Name (Employee MngrId=Id AND Salary>Salary Manager) The join yields a table with attributes: Employee.Name, Employee.Id, Employee.Salary, Employee.MngrId Manager.Name, Manager.Id, Manager.Salary Relational Algebra • Relational algebra operations operate on relations and produce relations (“closure”) f: Relation -> Relation Relation -> Relation • Six basic operations: – – – – – – Projection Selection Union Difference Product (Rename) A (R) (R) R1 [ R2 R1 – R2 R1 £ R2 A>B (R) f: Relation x Equijoin Join - Example Equijoin: Join condition is a conjunction of equalities. Name,CrsCode(Student Id=StudId Grade=‘A’ (Transcript)) Student Id 111 222 333 444 Transcript Name Addr Status StudId CrsCode Sem Grade John Mary Bill Joe ….. ….. ….. ….. ….. ….. ….. ….. 111 222 333 CSE305 S00 CSE306 S99 CSE304 F99 B A A Mary Bill CSE306 CSE304 The equijoin is used very frequently since it combines related data in different relations. Natural Join • Special case of equijoin + a special projection – join condition equates all and only those attributes with the same name (condition doesn’t have to be explicitly stated) – duplicate columns eliminated (projected out) from the result Transcript (StudId, CrsCode, Sem, Grade) Teaching (ProfId, CrsCode, Sem) Teaching = Transcript StudId, Transcript.CrsCode, Transcript.Sem, Grade, ProfId ( Transcript ) [StudId, CrsCode, Sem, Grade, ProfId ] CrsCode=CrsCode AND Sem=Sem Teaching Natural Join (cont’d) • More generally: R S = attr-list (join-cond (R × S) ) where attr-list = attributes (R) attributes (S) (duplicates are eliminated) and join-cond has the form: A1 = A1 AND … AND An = An where {A1 … An} = attributes(R) attributes(S) Natural Join Example • List all Ids of students who took at least two different courses: StudId ( CrsCode CrsCode2 ( Transcript Transcript [StudId, CrsCode2, Sem2, Grade2] )) We don’t want to join on CrsCode, Sem, and Grade attributes, hence renaming! Example Data Instance STUDENT Takes sid sid exp-grade name COURSE cid cid subj sem 1 Jill 1 A 550-0103 550-0103 DB F03 2 Qun 1 A 700-1003 700-1003 AI S03 3 Nitin 3 A 700-1003 501-0103 Arch F03 4 Marty 3 C 500-0103 4 C 500-0103 PROFESSOR Teaches fid name fid cid 1 Ives 1 550-0103 2 Saul 2 700-1003 8 Roth 8 501-0103 Natural Join and Intersection Natural join: special case of join where is implicit – attributes with same name must be equal: STUDENT ⋈ Takes ´ STUDENT ⋈STUDENT.sid = Takes.sid Takes Intersection: as with set operations, derivable from difference AB A-B A B-A B Division • A somewhat messy operation that can be expressed in terms of the operations we have already defined • Used to express queries such as “The fid's of faculty who have taught all subjects” • Paraphrased: “The fid’s of professors for which there does not exist a subject that they haven’t taught” Division Using Our Existing Operators • All possible teaching assignments: Allpairs: fid,subj (PROFESSOR £ subj(COURSE)) • NotTaught, all (fid,subj) pairs for which professor fid has not taught subj: Allpairs - fid,subj(Teaches ⋈ COURSE) • Answer is all faculty not in NotTaught: fid(PROFESSOR) - fid(NotTaught) ´ fid(PROFESSOR) - fid( fid,subj (PROFESSOR £ subj(COURSE)) fid,subj(Teaches ⋈ COURSE)) Division: R1 R2 • Requirement: schema(R1) ¾ schema(R2) • Result schema: schema(R1) – schema(R2) • “Professors who have taught all courses”: fid (fid,subj(Teaches ⋈ COURSE) subj(COURSE)) • What about “Courses that have been taught by all faculty”? Division • Goal: Produce the tuples in one relation, r, that match all tuples in another relation, s – r (A1, …An, B1, …Bm) – s (B1 …Bm) – r/s, with attributes A1, …An, is the set of all tuples <a> such that for every tuple <b> in s, <a,b> is in r • Can be expressed in terms of projection, set difference, and cross-product Division (cont’d) Division - Example • List the Ids of students who have passed all courses that were taught in spring 2000 • Numerator: – StudId and CrsCode for every course passed by every student: StudId, CrsCode (Grade ‘F’ (Transcript) ) • Denominator: – CrsCode of all courses taught in spring 2000 CrsCode (Semester=‘S2000’ (Teaching) ) • Result is numerator/denominator Relational Calculus • Important features: – Declarative formal query languages for relational model – Based on the branch mathematical logic known as predicate calculus – Two types of RC: • 1) tuple relational calculus • 2) domain relational calculus – A single statement can be used to perform a query Tuple Relational Calculus • based on specifying a number of tuple variables • a tuple variable refers to any tuple Generic Form • {t | COND (t)} – where – t is a tuple variable and – COND(t) is Boolean expression involving t Simple example 1 • To find all employees whose salary is greater than $50,000 – {t| EMPLOYEE(t) and t.Salary>5000} • where • EMPLOYEE(t) specifies the range of tuple variable t – The above operation selects all the attributes Simple example 2 • To find only the names of employees whose salary is greater than $50,000 – {t.FNAME, t.NAME| EMPLOYEE(t) and t.Salary>5000} • • • • The above is equivalent to SELECT T.FNAME, T.LNAME FROM EMPLOYEE T WHERE T.SALARY > 5000 Elements of a tuple calculus • In general, we need to specify the following in a tuple calculus expression: – Range Relation (I.e, R(t)) = FROM – Selected combination= WHERE – Requested attributes= SELECT More Example:Q0 • Retrieve the birthrate and address of the employee(s) whose name is ‘John B. Smith’ • {t.BDATE, t.ADDRESS| EMPLOYEE(t) AND t.FNAME=‘John’ AND t.MINIT=‘B” AND t.LNAME=‘Smith} Formal Specification of tuple Relational Calculus • A general format: • {t1.A1, t2.A2,…,tn.An |COND ( t1 ,t2 ,…, tn, tn+1, tn+2,…,tn+m)} – – – – where t1,…,tn+m are tuple var Ai : attributeR(ti) COND (formula) • Where COND corresponds to statement about the world, which can be True or False Elements of formula • A formula is made of Predicate Calculus atoms: – – – – – – an atom of the from R(ti) ti.A op tj.B op{=, <,>,..} F1 And F2 where F1 and F2 are formulas F1 OR F2 Not (F1) F’=(t) (F) or F’= (t) (F) • Y friends (Y, John) • X likes(X, ICE_CREAM) • Example Queries Using the Existential Quantifier • Retrieve the name and address of all employees who work for the ‘ Research ’ department • {t.FNAME, t.LNAME, t.ADDRESS| EMPLOYEE(t) AND ( d) (DEPARTMENT (d) AND d.DNAME=‘Research’ AND d.DNUMBER=t.DNO)} More Example • For every project located in ‘Stafford’, retrieve the project number, the controlling department number, and the last name, birthrate, and address of the manger of that department. Cont. • {p.PNUMBER,p.DNUM,m.LNAME,m.BD ATE, m.ADDRESS|PROJECT(p) and EMPLOYEE(M) and P.PLOCATION=‘Stafford’ and ( d) (DEPARTMENT(D) AND P.DNUM=d.DNUMBER and d.MGRSSN=m.SSN))} Safe Expressions • A safe expression R.C: – An expression that is guaranteed to generate a finite number of rows (tuples) • Example: – {t | not EMPLOYESS(t))} results values not being in its domain (I.e., EMPLOYEE) Examples • Retrieve the birthdates and address of the employee whose name is ‘John B. Smith’ • {uv| (q)(r)(s) (EMPLOYEE(qrstuvwxyz) and q=‘John’ and r=‘B’ and s=‘Smith’ Alternative notation • Ssign the constants ‘John’, ‘B’, and ‘Smith’ directly • {uv|EMPLOYEE (‘John’, ’B’, ’Smith’ ,t ,u ,v ,x ,y ,z)} More example • Retrieve the name and address of all employees who work for the ‘Reseach’ department • {qsv | ( z) EMPLOYEE(qrstuvwxyz) and ( l) ( m) (DEPARTMENT (lmno) and l=‘Research’ and m=z))} More example • List the names of managers who have at least on e dependent • {sq| ( t) EMPLOYEE(qrstuvwxyz) and (( j)( DEPARTMENT (hijk) and (( l) | (DEPENTENT (lmnop) and t=j and t=l))))} Relational Calculus: A Logical Way of Expressing Query Operations • First-order logic (FOL) can also be thought of as a query language, and can be used in two ways: – Tuple relational calculus – Domain relational calculus – Difference is the level at which variables are used: for attributes (domains) or for tuples • The calculus is non-procedural (declarative) as compared to the algebra – More like what we’ll see in SQL – More convenient to express certain things More Complex Predicates Starting with these atomic predicates, build up new predicates by the following rules: – Logical connectives: If p and q are predicates, then so are pq, pq, p, and pq • (x>2) (x<4) • (x>2) (x>0) – Existential quantification: If p is a predicate, then so is x.p • x. (x>2) (x<4) – Universal quantification: If p is a predicate, then so is x.p • x.x>2 • x. y.y>x Logical Equivalences • There are two logical equivalences that will be heavily used: – pq p q (Whenever p is true, q must also be true.) – x. p(x) x. p(x) (p is true for all x) • The second can be a lot easier to check! Normalization Review on Keys • • • • • • superkey: a set of attributes which will uniquely identify each tuple in a relation candidate key: a minimal superkey primary key: a chosen candidate key secondary key: all the rest of candiate keys prime attribute: an attribute that is a part of a candidate key (key column) nonprime attribute: a nonkey column Normalization Functional Dependency Type by Keys ‘whole (candidate) key nonprime attribute’: full FD (no violation) • ‘partial key nonprime attribute’: partial FD (violation of 2NF) • ‘nonprime attribute nonprime attribute’: transitive FD (violation of 3NF) • ‘not a whole key prime attribute’: violation of BCNF • Perform lossless-join decompositions of each of the following scheme into BCNF schemes: R(A, B, C, D, E) with dependency set {AB CDE, C D, D E} A B C D C D A B A B C E D E A B C D C D A B C D E C D A B C Given the FDs {B D, AB C, D B} and the relation {A, B, C, D}, give a two distinct lossless join decomposition to BNCF indicating the keys of each of the resulting relations. A B B D C D A B A B C B D C D A C D Definition of MVD • A multivalued dependency (MVD) X ->->Y is an assertion that if two tuples of a relation agree on all the attributes of X, then their components in the set of attributes Y may be swapped, and the result will be two tuples that are also in the relation. Example • The name-addr-phones-beersLiked example illustrated the MVD name->->phones and the MVD name ->-> beersLiked. Picture of MVD X ->->Y X Y equal exchange others MVD Rules • Every FD is an MVD. – If X ->Y, then swapping Y ’s between two tuples that agree on X doesn’t change the tuples. – Therefore, the “new” tuples are surely in the relation, and we know X ->->Y. • Complementation : If X ->->Y, and Z is all the other attributes, then X ->->Z. Fourth Normal Form • The redundancy that comes from MVD’s is not removable by putting the database schema in BCNF. • There is a stronger normal form, called 4NF, that (intuitively) treats MVD’s as FD’s when it comes to decomposition, but not when determining keys of the relation. 4NF Definition • A relation R is in 4NF if whenever X ->->Y is a nontrivial MVD, then X is a superkey. – “Nontrivial means that: 1. Y is not a subset of X, and 2. X and Y are not, together, all the attributes. – Note that the definition of “superkey” still depends on FD’s only. BCNF Versus 4NF • Remember that every FD X ->Y is also an MVD, X ->->Y. • Thus, if R is in 4NF, it is certainly in BCNF. – Because any BCNF violation is a 4NF violation. • But R could be in BCNF and not 4NF, because MVD’s are “invisible” to BCNF. Normalization Good Decomposition • dependency preserving decomposition - it is undesirable to lose functional dependencies during decomposition • lossless join decomposition - join of decomposed relations should be able to create the original relation (no spurious tuples) Decomposition and 4NF • If X ->->Y is a 4NF violation for relation R, we can decompose R using the same technique as for BCNF. 1. XY is one of the decomposed relations. 2. All but Y – X is the other. Example Drinkers(name, addr, phones, beersLiked) FD: MVD’s: name -> addr name ->-> phones name ->-> beersLiked • Key is {name, phones, beersLiked}. • All dependencies violate 4NF. Example, Continued • Decompose using name -> addr: 1. Drinkers1(name, addr) In 4NF, only dependency is name -> addr. 2. Drinkers2(name, phones, beersLiked) Not in 4NF. MVD’s name ->-> phones and name ->-> beersLiked apply. No FD’s, so all three attributes form the key. Example: Decompose Drinkers2 • Either MVD name ->-> phones or name ->-> beersLiked tells us to decompose to: – Drinkers3(name, phones) – Drinkers4(name, beersLiked)

Descargar
# Document