Plan of the lecture. G53RDB: Theory of Relational Databases Lecture 2. More operations: renaming. Previous lecture. Renaming.

Transcription

1 Plan of the lecture G53RDB: Theory of Relational Lecture 2 Natasha Alechina chool of Computer cience & IT nza@cs.nott.ac.uk Renaming Joins Definability of intersection Division ome properties of relational algebra operators Exercises Lecture 2 2 Previous lecture More operations: renaming Let R, be union compatible, that is R A 1... A n and A 1... A n. Then R = {a A 1... A n : a R or a } (union) R = {a A 1... A n : a R and a } (intersection) R = {a A 1... A n : a R and a } (difference) R = {< a 1,, a n, b 1,, b m >: < a 1,, a n > R and < b 1,, b m > } (product) π X (R) = {a(x): a R} (projection) σ α (R) = {<a 1,,a n > R: α (a 1,,a n )} (selection) Each attribute in a schema should be unique but this is not guaranteed when taking a product of two relations. R Name Age Joe 20 Jill 21 Name Address Tom Bill R NameAge NameAddress Joe 20 Tom Joe 20 Bill Jill 21 Tom Jill 21 Bill Lecture 2 3 Lecture 2 4 Renaming Equi-join An operation to rename a relation R to be and to rename attributes of R to be A 1,,A n (in order from left: first attribute of R is now called A 1, etc.): ρ (A1,,An ) (R). The resulting relation has exactly the same set of tuples, but a different name and a different schema. Note that if we only want to rename one of the attributes, we can do this with this operation. For example, let the old schema of R be {A,B,C} and the desired schema of {A,B,E}: ρ (A,B,E) (R). To just rename the relation R to, we will write ρ (R). We define this operation as an intermediate step to defining natural join. Equi-join is an operation which takes two relations R and and two equality-comparable attributes: Att i in the schema of R and Att j in the schema of, and produces a new relation which contains tuples {<a 1,,a n,b 1,,b m >: <a 1,,a n > R, <b 1,,b m >, and a i = b j } Lecture 2 5 Lecture 2 6 1

2 Equi-join: example Natural join R R equi-join NR = N NR Tel N Addr NR Tel N Addr Natural join >< is equi-join with the duplicate attribute removed (note that the values in two columns for Att i and Att j are always identical). Joe 1234 Joe Joe 1234 Joe Jill 1244 Jill Jill 1244 Jill Bill 2244 Bob NG1 Lecture 2 7 Lecture 2 8 Natural join: example Definition of natural join R R >< NR = N NR Tel N Addr NR Tel Joe 1234 Joe Joe 1234 Jill 1244 Jill Jill 1244 Addr Given a relation R over schema which contains attribute Att i and relation over schema which contains attribute Att j, so that Att i and Att j are =-comparable, R >< i = j = {<a 1,,a n,b 1,,b j-1, b j+1,, b m >: <a 1,,a n > R, <b 1,,b m >, and a i = b j } Bill 2244 Bob NG1 Lecture 2 9 Lecture 2 10 Natural join is definable Intersection is definable Natural join is a very useful operator, but it can be defined using, π and σ. To define equi-join of R and over attributes X and Y we take a Cartesian product of R and and then select tuples which satisfy X = Y. To obtain natural join, we project the resulting relation on the set of all attributes apart from Y. Let A be the set of all attributes in R and : R >< X =Y = π A - Y σ X = Y (R ) R = R (R ) R R Lecture 2 11 Lecture

3 Intersection is definable Intersection is definable R = R (R ) R R = R (R ) R R R (R ) Lecture 2 13 Lecture 2 14 Division Let R be a relation over schema {X 1,,X n, Y 1,,Y m }. We will denote {X 1,,X n } by X and {Y 1,,Y m } by Y. Let be a relation with schema Y. Then R ={a: a π X (R) and for all b, a ~b R} (if a = <a 1,,a k > and b= <b 1,,b m >, then a concatenated with b, a ~ b = < a 1,,a k, b 1,,b m >) R R Lecturer Module ubject Lecturer mith Lecture 2 15 Lecture 2 16 R R Lecturer Module ubject Lecturer R R Lecturer Module ubject Lecturer mith mith Lecture 2 17 Lecture

4 Defining R Defining R Consider the relation π X (R). It contains all tuples a~b starting with a tuple a from π X (R) and ending in a tuple from b from. Take any tuple a R. For all b, a~b is in R by definition of division. On the other hand, for any tuple a π X (R) which is not in R, there is a b, such that a~b is not in R but it is in π X (R). o let s look at (π X (R) ) R. It will contain only tuples a~b where a R. Hence, π X ((π X (R) ) R) is the set of a π X (R) which are not in R. Finally, to define R, we need to subtract the previous set from π X (R): R = π X (R) π X ((π X (R) ) R) Lecture 2 19 Lecture 2 20 Relational algebra operators We defined operations on relations which take relations as operands and return a relation. By relational algebra we will mean the following set of operators: (,,, π, σ, ρ) They are not functionally complete in the sense that boolean algebra operators are (e.g. AND and NOT are sufficient to express any boolean-valued function). ome operations on relations are not expressible in terms of the operations we have (e.g. transitive closure of a relation.) Next we look at the algebraic properties of our operations (we need them because we are going to use them in query rewriting). ome properties of Commutativity: R = R Associativity: R (P ) = (R P ) If R then R = Lecture 2 21 Lecture 2 22 ome properties of Commutativity: R = R Associativity: R (P ) = (R P ) If R then R = R ome properties of and Distributivity: R (P ) = (R P) (R ) R (P ) = (R P) (R ) Lecture 2 23 Lecture

5 ome properties of difference ome properties of R (P ) = (R P) (R ) R (P ) = (R P) (R ) R R = R = R Associativity: R (P ) = (R P) ) Distributivity: R (P ) = (R P) (R ) R (P ) = (R P) (R ) R = R = Lecture 2 25 Lecture 2 26 Exercises Recommended reading Prove distributivity of Cartesian product over union: R (P ) = (R P) (R ) using definitions of union and product. Does selection satisfy the following properties: σ α (σ α (R)) = σ α ( R) σ β (σ α (R)) = σ α&β ( R) σ β (σ α (R)) = σ α (σ β ( R)) Does projection satisfy the following properties: π X ( π X (R)) = π X (R) π X ( π Y (R)) = π Y ( π X (R)) Ullman, Widom: Chapter 5.1, 5.2. Lecture 2 27 Lecture