1 / 58

RELATIONAL ALGEBRA (II)

CS157A. Lecture 9. RELATIONAL ALGEBRA (II). Prof. Sin-Min LEE Department of Computer Science. Unary Relational Operations: SELECT and PROJECT. The PROJECT Operation Sequences of Operations and the RENAME Operation The SELECT Operation. Relational Algebra Operations from Set Theory.

kirkan
Download Presentation

RELATIONAL ALGEBRA (II)

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. CS157A Lecture 9 RELATIONAL ALGEBRA (II) Prof. Sin-Min LEE Department of Computer Science

  2. Unary Relational Operations: SELECT and PROJECT • The PROJECT Operation • Sequences of Operations and the RENAME Operation • The SELECT Operation

  3. Relational Algebra Operations from Set Theory • The UNION, INTERSECTION, and MINUS Operations

  4. Binary Relational Operations: JOIN and DIVISION • The JOIN Operation • The EQUIJOIN and NATURAL JOIN Variations of JOIN • A Complete Set of Relational Algebra Operations • The DIVISION Operation

  5. Additional Relational Operations • Aggregate Functions and Grouping • Recursive Closure Operations • OUTER JOIN Operations • The OUTER JOIN Operation

  6. SPECIAL RELATIONAL OPERATORS The following operators are peculiar to relations: - Join operators There are several kind of join operators. We only consider three of these here (others will be considered when we discuss null values): - (1) Condition Joins - (2) Equijoins - (3) Natural Joins - Division

  7. JOIN OPERATORS Condition Joins: - Defined as a cross-product followed by a selection: R ⋈cS = σc(R  S) (⋈ is called the bow-tie) where c is the condition. - Example: Given the sample relational instances S1 and R1 The condition join S ⋈S1.sid<R1.sid R1 yields

  8. Equijoin: Special case of the condition join where the join condition consists solely of equalities between two fields in R and S connected by the logical AND operator (∧). Example: Given the two sample relational instances S1 and R1 The operator S1 R.sid=Ssid R1 yields

  9. Natural Join - Special case of equijoin where equalities are implicitly specified on all fields having the same name in R and S. - The condition c is now left out, so that the “bow tie” operator by itself signifies a natural join. - N. B. If the two relations have no attributes in common, the natural join is simply the cross-product.

  10. DIVISION - The division operator is used for queries which involve the ‘all’ qualifier such as “Find the names of sailors who have reserved all boats”. - The division operator is a bit tricky to explain.

  11. EXAMPLES OF DIVISION

  12. DIVISION Example: Find the names of sailors who have reserved all boats: (1) A = sid,bid(Reserves). A1 = sid(Reserves) A2 = bid(Reserves) (2) B2 = bid(Boats) B3 is the rest of B. Thus, B2 ={101, 102, 103, 104} (3) Find the rows of A such that their A.sid is the same and their combined A.bid is the set B2. Thus we find A1 = {22} (4) Get the set of A2 corresponding to A1: A2 = {Dustin}

  13. FORMAL DEFINITION OF DIVISION The formal definition of division is as follows: A/B = x(A) - x((x(A)  B) – A)

  14. EXAMPLES OF ALGEBRA QUERIES In the rest of this chapter we shall illustrate queries using the following new instances S3 of sailors, R2 of Reserves and B1 of boats.

  15. QUERY Q1 Given the relational instances: (Q1) Find the names of sailors who have reserved boat 103 sname((σbid=103Reserves) ⋈ Sailors) The answer is thus the following relational instance {<Dustin>, <Lubber>, <Horatio>}

  16. QUERY Q1 (cont’d) There are of course several ways to express Q1 in relational algebra. Here is another: sname(σbid=103(Reserves⋈ Sailors)) Which of these expressions should we use? That is a question of optimization. Indeed, when we describe how to state queries in SQL, we can leave it to the optimizer in the DBMS to select the nest approach.

  17. QUERY Q2 (Q2) Find the names of sailors who have reserved a red boat. sname((σcolor=‘red’Boats) ⋈ Reserves ⋈ Sailors)

  18. QUERY Q3 (Q3) Find the colors of boats reserved by Lubber. color((σsname=‘Lubber’Sailors)Sailors ⋈ Reserves ⋈ Boats)

  19. QUERY Q4 (Q4) Find the names of Sailors who have reserved at least one boat sname(Sailors ⋈ Reserves)

  20. QUERY Q5 (Q5) Find the names of sailors who have reserved a red or a green boat. (Tempboats, (σcolor=‘red’Boats) ∪ (σcolor=‘green’Boats)) sname(Tempboats⋈Reserves⋈Sailors)

  21. QUERY Q6 (Q6) Find the names of Sailors who have reserved a red and a green boat. It seems tempting to use the expression used in Q5, replacing simply ∪ by ∩. However, this won’t work, for such an expression is requesting the names of sailors who have requested a boat that is both red and green! The correct expression is as follows: (Tempred, sid((σcolor=‘red’Boats) ⋈Reserves)) (Tempgreen, sid((σcolor=‘green’Boats) ⋈Reserves)) sname ((Tempred∩Tempgreen) ⋈ Sailors)

  22. QUERY Q7 (Q7) Find the names of sailors who have reserved at least two boats. (Reservations, sid,sname,bid(Sailors ⋈Reserves)) (Reservationpairs(1sid1, 2sname, 3bid1, 4sid2, 5sname, 6bid2), ReservationsReservations) sname1σ(sid1=sid2)(bid1bid2)Reservationpairs)

  23. QUERY 8 (Q8) Find the sids of sailors with age over 20 who have not reserved a red boat. sid(σage>20Sailors) - sid((σcolor=‘red’Boats)⋈Reserves⋈Sailors)

  24. QUERY 9 (Q) Find the names of sailors who have reserved all boats. (Tempsids, (sid,bidReserves) / (bidBoats)) sname(Tempsids⋈Sailors

  25. QUERY Q10 (Q10) Find the names of sailors who have reserved all boats called Interlake. (Tempsids, (sid,bidReserves)/(bid(σbname=‘Interlake’Boats))) sname(Tempsids⋈Sailors)

  26. Cartesian Product • R(A1, A2, ..., Am) and S(B1, B2, ... , Bn) • T(A1, A2, ... , Am, B1, B2, ..., Bn) = R(A1, A2, ..., Am) X S(B1, B2, ..., Bn) • A tuple t is in T if and only if t[A1,A2, ... ,Am] is in R and t[B1, B2, ..., Bn] is in S. - If R has N1 tuples and S has N2 tuples, then T will have N1*N2 tuples.

  27. Cartesian Product R Rx S S

  28. Examples of Queries in Relational Algebra

  29. Natural-Join • Denoted by |x|. • Binary operation • Creates a Cartesian-product of the arguments then performs selection to force equality on attributes that appear in both relations

  30. Division • Denoted by  • Binary Operation • Used in queries that include the phrase “for all”.

  31. Division (Cont’d) • Division is an operation on schema R – S • A tuple t is in r  s if and only if: • t is in ΠR – S(r) and • For every tuple ts in s, there is a tuple tr in r satisfying both of the following: a. tr[S] = ts[R] b. tr[R – S] = t

  32. Relational Algebra Fundamental operators • select s • project p • cartesian product  • union  • set difference - Other operators • natural join JOIN (butterfly symbol) • set intersection  • division 

  33. A Simple DB account ac# owner ss# balance 1 bob 123 1000 2 sue 456 2000 3 jane 789 3000 transaction t# ac# type amount outcome date 1 1 W 1500 bounced 5/1/98 2 2 D 1000 ok 5/2/98 3 1 W 100 ok 5/4/98 4 3 D 500 ok 5/7/98 5 2 W 200 ok 5/9/98 account had transaction

  34. Select eg: s balance>=1500 account result : ac# owner ss# balance 2 sue 456 2000 3 jane 789 3000 Project eg: π owner, ss# account result: owner ss# bob 123 sue 456 jane 789

More Related