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RELATIONAL ALGEBRA (III)

CS157A. RELATIONAL ALGEBRA (III). Prof. Sin-Min LEE Department of Computer Science. Lecture 10. 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.

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RELATIONAL ALGEBRA (III)

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  1. CS157A RELATIONAL ALGEBRA (III) Prof. Sin-Min LEE Department of Computer Science Lecture 10

  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 • The CARTESIAN PRODUCT (or CROSS PRODUCT) Operation

  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. 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

  9. 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

  10. 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.

  11. 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, and perhaps best approached through examples as will be done here.

  12. EXAMPLES OF DIVISION

  13. DIVISION Interpretation of the division operation A/B: - Divide the attributes of A into 2 sets: A1 and A2. - Divide the attributes of B into 2 sets: B2 and B3. - Where the sets A2 and B2 have the same attributes. - For each set of values in B2: - Search in A2 for the sets of rows (having the same A1 values) whose A2 values (taken together) form a set which is the same as the set of B2’s. - For all the set of rows in A which satisfy the above search, pick out their A1 values and put them in the answer.

  14. 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}

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

  16. 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.

  17. 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>}

  18. 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.

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

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

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

  22. 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)

  23. 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)

  24. 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)

  25. 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)

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

  27. 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)

  28. Natural Join - combines σ, π,  - very commonly used Natural Join forms the cross product of its two arguments, does a selection to enforce equality of columns with the same name and removes duplicate columns. Eg: “show all transactions done by account owner Bob” σ owner=“Bob”(account JOIN transaction)

  29. Rename operation What if you need to access the same relation twice in a query? eg. person(ss#, name, mother_ss#, father_ss#) “Find the name of Bob’s mother” needs the “person” table to be accessed twice. The operation ρ x (r) evaluates to a second logical copy of relation r renamed to x.

  30. Rename operation (contd) eg: π mother.name ( (ρ mother (person)) JOIN mother.ss# = person.mother_ss# (sname=“Bob” (person)))

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