Query Processing. 3 steps: Parsing & Translation Optimization Evaluation

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Laurence Mosley
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1 rela%onal algebra

2 Query Processing 3 steps: Parsing & Translation Optimization Evaluation 30

3 Simple set of algebraic operations on relations Journey of a query SQL select from where Rela%onal algebra π 13 (P Q) Query rewri%ng π 14 (P S) Q R We use set seman/cs (no duplicates) and no nulls There are extensions with bag seman%cs and nulls 31

4 Projection Eliminate some columns π X (R) Display only attributes X of relation R where R: table name & X attributes(r) Example: Find titles of current movies π TITLE (SCHEDULE) 32

5 Projection Eliminate some columns π X (R) Display only attributes X of relation R where R: table name & X attributes(r) Example: R A B C π A (R) = A π AB (R) = A B No repetitions of tuples! 33

6 Selection Compute set union σ cond (R) Select tuples of R satisfying condition cond where cond: condition involving only attributes of R (e.g., attr = value, attr value, attr1 = attr2, attr1 attr2, etc.) Example: R A B C A B C σ Α=0 (R) = σ B=C (R) = A B C

7 Selection Compute set union σ cond (R) Select tuples of R satisfying condition cond where cond: condition involving only attributes of R (e.g., attr = value, attr value, attr1 = attr2, attr1 attr2, etc.) Example: R A B C A B C σ Α 0 (R) =

8 Union Compute set union R S Union of sets of tuples in R and S where R, S: tables with same attributes Example: R A B S A B R S = A B α 1 α 2 α 1 α 2 β 3 α 2 β 1 β 1 β 3 36

9 Difference Compute set difference R - S Difference of sets of tuples in R and S where R, S: tables with same attributes Example: R A B S A B R - S = A B α 1 α 2 α 1 α 2 β 3 β 1 β 1 37

10 Join Compute join R S Natural Join of R, S where R, S: tables Example: R A B S B C R S = A B C Note: More than one common attributes allowed! 38

11 Join Compute join R S Natural Join of R, S where R, S: tables Example: R A B S B C R S = A B C

12 Definition of Join Let r and s be relations on schemas R and S respectively. Then, r s is a relation with attributes att(r) att(s) obtained as follows: Consider each pair of tuples t r from r and t s from s. If t r and t s have the same value on each of the attributes in att(r) att(s), add a tuple t to the result, where t has the same value as t r on r t has the same value as t s on s Note: if R S is empty, the join consists of all combinations of tuples from R and S, i.e. their cross-product

13 Attribute Renaming Rename attributes δ A1 A2 (R) Change name of attribute A1 in rel. R to A2 where R: relation and A1: attribute in R Example: R A B δ A C (R) = C B α 1 α 1 α 2 α 2 β 1 β 1 Contents remain unchanged! Note: Can rename several attributes at once 41

14 Basic set of operations: π, σ,, -,, δ Back to movie example queries: 1. Titles of currently playing movies: π TITLE (schedule) 2. Titles of movies by Berto: π TITLE (σ DIR=BERTO (movie)) 3. Titles and directors of currently playing movies: π TITLE, DIR (movie schedule) 42

15 4. Find the pairs of actors acting together in some movie π actor1, actor2 (δ actor à actor1 (movie) δ actor à actor2 (movie)) 5. Find the actors playing in every movie by Berto π actor (movie) π actor [(π actor (movie) π title (σ dir =BERTO (movie))) - π actor,title (movie)] actor %tle by Berto actor acts in %tle actors for which there is a movie by Berto in which they do not act In this case (not in general): Same as cartesian product 43

16 Cartesian Product Compute cartesian product R S Cartesian Product of R, S where R, S: tables Example: R A B S C D R S = A B C D Same as R S, when R and S have no common attributes

17 Cartesian Product Compute cartesian product R S Cartesian Product of R, S where R, S: tables Example: R A B S A C R S = R.A B S.A C If 2 attributes in R, S have the same name A, they are renamed to R.A and S.A in the output

18 Other useful operations Intersection R S Division (Quotient) R S R A B S B R S: {a <a, b> R for every b S} Example: R A B S B R S = 0 α α 0 β β 1 α 1 β 1 γ 2 α Α

19 Another Division Example Find the actors playing in every movie by Berto" π TITLE, ACTOR (movie) π TITLE (σ DIR=BERTO (movie))

20 Division by multiple attributes Relations r, s:" r! A! B! C! D! E! s! D! E! α! α! α! β! β! γ! γ! γ! a! α! γ! γ! γ! γ! γ! γ! β! b" b" b" b! 1! 1! 1! 1! 3! 1! 1! 1! b! 1! 1! r s:" A! B! C! α! γ! a! γ! γ!

21 Note: π is like there exists is like for all Expressing using other operators: R S = π A (R) - π A ((π A (R) S) - R) R A B S B Similar to: x ϕ(x) x ϕ(x) 49

22 Join Operators Compute different types of joins R S R S R S R S R θ S Natural Join Left Outer Join Right Outer Join Full Outer Join Theta Join: Same as σ θ (R S) 50

23 Calculus Vs. Algebra Theorem: Calculus and Algebra are equivalent Basic Correspondence: Algebra Operation π σ - Calculus Operation t(a) comp c 51

24 Example Find theaters showing movies by Bertolucci : SQL: SELECT s.theater FROM schedule s, movie m WHERE s.title = m.title AND m.director = Berto tuple calculus: { t: theater s schedule m movie [ t(theater) = s(theater) s(title) = m(title) m(director) = Berto ] } relational algebra: π theater (schedule σ dir = Berto (movie)) Note: number of items in FROM clause = (number of joins + 1) 52

Computational Logic. Relational Query Languages with Negation. Free University of Bozen-Bolzano, Werner Nutt

Computational Logic. Relational Query Languages with Negation. Free University of Bozen-Bolzano, Werner Nutt Computational Logic Free University of Bozen-Bolzano, 2010 Werner Nutt (Slides adapted from Thomas Eiter and Leonid Libkin) Computational Logic 1 Queries with All Who are the directors whose movies are