Relational Databases. Relational Databases. Extended Functional view of Information Manager. e.g. Schema & Example instance of student Relation

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1 Relational Databases Relational Databases 1 Relational Model of Data 2 Relational Algebra (and connection to Datalog) Relational database: a set of relations/tables Relation schema : specifies name of relation, fixed set of named columns (a.k.a. fields, attributes) additional information/constraints about the columns E.G. student(sid, Name, Age, Gpa). (Note: in logic/datalog, columns are identified by position, not name). Relation instance: is a set of tuples (a.k.a. rows, records) of values in the columns. Think of it as a table (As a set, it is not supposed to have duplicates! ) Must match the schema 336 S14 A.Borgida S14 A.Borgida 2 Schema & Example instance of student Relation student(sid,name,age,gpa) sid name age gpa 5366 Jones Smith Smith v cardinality (number of rows)= 3, all rows distinct v aritiy (number of columns) =4 Note that the schema is (mostly) stable/unchanging, and the instance may change frequently, reflecting the state of the world. L tell Extended Functional view of Information Manager TELL Info Manager TELL: L tell IM IM È Exceptions ASK: L question IM L answer DEFINE: L declare IM IM L declare/constrain DECLARE / CONSTRAIN ASK L question L answer 336 S14 A.Borgida S14 A.Borgida 4

2 Relational databases L declare/constrain : SQL syntax for tables CREATE TABLE student (sid INT, name VARCHAR, grade CHAR(2), gpa REAL, PRIMARY KEY (sid) ) CREATE TABLE enrolledin (sid INT, cid INT, grade CHAR(2), PRIMARY KEY (sid,cid) ) A set of fields is a key if no distinct tuples can have same key values. No subset of these columns should have this property. This does not allow retaking a class! Relational databases L tell : ground atomic formulae as tuples in tables (no rules!) Datalog: likes(bob,eve). SQL: INSERT INTO likes VALUES ( bob, eve ); L question : 1. relational algebra 2. tuple calculus : SQL 3. domain calculus : arbitrary First Order Logic formulae with free variables (Datalog + universal quantifier) L answer : set of tuples in a table 336 S14 A.Borgida S14 A.Borgida 7 Relational Algebra What is an Algebra? constants expressions made of operators applied to operand expressions laws for manipulating expressions to get equivalent ones, Algebra of sets: start from sets defined by enumeration operators: union, intersection, complement of Relational Tables not really used for querying by end-users more like instructions to compute answers but useful terminology (everyone uses it), and used in making queries run more efficiently (optimization) Table name student Constants Find all students ( Datalog answer(sid,name,age,gpa) :- student(sid,name,age,gpa) 336 S14 A.Borgida S14 A.Borgida 9

3 Eliminates columns Project operator Find name and age of students PROJECT[#2,#3](student) PROJECT[Name,Age](student) Student p #2,#3 (student) p Name,Age (student) PROJECT[name,gpa](Student) ( Datalog answer(name,age) :- student(_,name,age,_) 336 S14 A.Borgida 10 (eliminated because duplicate) 336 S14 A.Borgida 11 Select operator Eliminates rows Find all information about students whose age = 18 SELECT[Age=18](student) s Age=18 (student) << ascii notation << math notation answer(sid,name,age,year) :- student(sid,name,age,year), Age =18 sid name age gpa 5366 Jones Smith Smith SELECT[Age=18](student) sid name age gpa 5366 Jones Smith original Student table answer 336 S14 A.Borgida S14 A.Borgida 13

4 Algebra: combine things Find name of 18 year old students PROJECT[name] ( SELECT[age=18] (student) ) Product operator Combines relations (cartesian product in math) Find all possible combinations of enrolledin and course (not very useful on its own) PRODUCT(enrolledIn,course) enrolledin course Name conflict needs to be resolved answer(sid,enrolledin_cid,grade,course_cid,title,dept) :- enrolledin(sid,enrolledin_cid,grade), course(course_cid,title,dept). 336 S14 A.Borgida S14 A.Borgida 15 R T R a1 p a2 q b1 b2 q s t c1 d1 e1 c2 d2 e2 c3 d3 e3 p R.q T.q s t a1 b1 c1 d1 e1 a1 b1 c2 d2 e2... a2 b2 c3 d3 e3 We will assume that when taking the product of two different tables, the renaming of conflicting column names is automatic, as above. Note that this will not work for R R!! Here we need RENAME[ ] 336 S14 A.Borgida 16 T Union, Intersect Combines relations using set-operations; must have compatible schema (same column names and datatypes) Find Sid of young students or A students student1 := PROJECT[Sid]( SELECT[Age<18](students) ) student2 := PROJECT[Sid]( SELECT[grade= A ](enrolledin)) s3 := UNION(student1,student2) (note the use of named intermediate relations to clarify steps) p Sid ( s Age=18 (student)) p Sid ( s Grade= A (enrolledin)) answer(sid):- student(sid,name,age,year), Age < 18. answer(sid):- enrolledin(sid,_, A ). 336 S14 A.Borgida 17

5 Students not taking any course Difference also set theoretic DIFF( PROJECT[Sid](student), PROJECT[Sid](enrolledIn)) p Sid (student) - p Sid (enrolledin) takesomecrs(sid):- enrolledin(sid,_,_,_). student_sid(sid) :- student(sid,_,_,_). answer(sid):- student_sid(sid), NOT takesomecrs(sid) (What if you wanted their name too?) 336 S14 A.Borgida 18 OR Rename table and/or column (Trivial name change. Necessary when combination of multiple tables, esp the same table with itself, results in conflicting names; or when doing set ops) Compute a table 'taking' of 'scores' gotten by students. RENAME[taking(Grade~> Score)] (enrolledin) taking := RENAME[Grade~> Score] (enrolledin) r taking(grade~> Score) (enrolledin) taking(sid,cid,score):-,grade=score. 336 S14 A.Borgida 19 Try Pairs of student Ids, taking the same course Natural Join (very useful) Get all students with their courses and grades JOIN(student,enrolledIn) student >< enrolledin Resulting schema: set of attributes in both schemas R and S (no duplicates) Tuples: best described by either the Datalog below of equivalence on next slide answer(sid,name,age,year,cid,grade):- student(sid,name,age,year), 336 S14 A.Borgida S14 A.Borgida 21

6 Algebraic Identity JOIN(student,enrolledIn) = PROJECT[Sid,Name,Age,Year,Cid,Grade] (SELECT[Sid=Sid_t] (PRODUCT(student, RENAME[Sid~>Sid_t](enrolledIn)))) 336 S14 A.Borgida S14 A.Borgida 23 Generalized Join: Theta Join Find courses in which a student got less than their GPA THETA_JOIN[student.sid=enrolled.sid & gpa>grade] (student,enrolledin) Following examples are from textbook authors and are not covered in class. student >< Condition enrolledin Same as SELECT[Condition] (PRODUCT(student, enrolledin)) 336 S14 A.Borgida S14 A.Borgida 27

7 Find names of sailors who ve reserved boat #103 Find names of sailors who ve reserved a red boat Solution 1: π sname ((σ bid=103 Reserves) Sailors) Information about boat color only available in Boats; so need an extra join: Solution 2: π sname ((σ color='red' Boats) Reserves Sailors) π sname ( Temp2) v A more efficient solution: π sname (π sid ((π bid σ color ='red' Boats) Res) Sailors) Solution 3: π sname (σ bid=103 (Reserves Sailors)) A query optimizer can find this, given the first solution! 336 S14 A.Borgida S14 A.Borgida 29 Find sailors who ve reserved a red or a green boat Can identify all red or green boats, then find sailors who ve reserved one of these boats: ρ (Tempboats,(σ color ='red' color ='green' Boats)) Find sailors who ve reserved a red and a green boat Previous approach won t work! Must identify sailors who ve reserved red boats, sailors who ve reserved green boats, then find the intersection (note that sid is a key for Sailors): π sname (Tempboats Reserves Sailors) v Can also define Tempboats using union! (How?) ρ (Tempred,π sid ((σ color ='red' Boats) Reserves)) ρ (Tempgreen,π sid ((σ color ='green' Boats) Reserves)) π sname ((Tempred Tempgreen) Sailors) 336 S14 A.Borgida S14 A.Borgida 31