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Database System Concepts, 5th Ed. ©Silberschatz, Korth and Sudarshan See www.db-book.com for conditions on re-usewww.db-book.com ICOM 5016 – Database Systems Dr. Amir H. Chinaei Department of Electrical and Computer Engineering University of Puerto Rico, Mayagüez Slides are adapted from: Relational Algebra

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2 Fundamental Property Operands: relations Operators: the most required functions in relational model Unary or Binary Result is also a relation Introduction. one of the two formal query languages associated with the relational model.

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3 Roadmap Core R.A. includes basic operators Advanced R.A. includes additional operators (that mostly can be defined in terms of Core R.A.) Extended R.A. Extends R.A. to support more features of SQL

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4 R.A. Operators Basic Selection Projection Union Difference (Cross-)product, Additional Intersection Rename Joins Division

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5 Selection Selection extracts some rows from a relation R2 ::= C (R1) R2 includes only those rows of R1 that satisfy C C is a Boolean condition

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6 Example: Selection FoodserviceFoodPrice Brubaker’s Bon Appetite Sushi Stroganoff Meatball Sushi Meatball 7.50 6.00 5.00 8.00 5.00 Provides R2 ::= price<6.50 (Provides) FoodserviceFoodPrice Brubaker’s Bon Appetite Stroganoff Meatball 6.00 5.00 R2

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7 Projection Projection extracts some columns from a relation R2 := A (R1) A consists of some attributes of R1 R2 includes only those columns of R1 that are in A. Eliminate duplicate tuples, if any.

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8 Example: Projection FoodserviceFoodPrice Brubaker’s Bon Appetite Sushi Stroganoff Meatball Sushi Meatball 7.50 6.00 5.00 8.00 5.00 Provides R2 ::= Foodservice, Food (Provides) R2 FoodserviceFood Brubaker’s Bon Appetite Sushi Stroganoff Meatball Sushi Meatball

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9 Set-Based Operators R.A. borrows the following standard operators on sets: Union. R3 ::= R1 U R2 Intersection. R3 ::= R1 ∩ R2 Difference. R3 ::= R1 - R2 Cross-Product (Cartesian product). R3 ::= R1 R2

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10 Union Union puts together all tuples that are in R1 or R2 or both R3 ::= R1 U R2 R1 and R2 must be union-compatible: They have the same number of attributes Corresponding attributes have the same domains

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11 Intersection Intersection puts together all tuples that are in both relations R1 and R2 R3 ::= R1 ∩ R2 R1 and R2 must be union-compatible

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12 Difference Difference consists of those tuples of first relation which do not appear in the other R3 ::= R1 - R2 R1 and R2 must be union-compatible

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13 Product Product pairs each tuple t1 of R1 with each tuple t2 of R2 R3 ::= R1 R2 Schema of R3 is the attributes of R1 and R2, in order If an attribute name A is repeated in both R1 and R2, it is represented by R1.A and R2.A

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14 Example: Product FSFP a1 a2 a3 b1 b2 b3 c1 c2 c3 NF d1 d2 b1 b2 R1 R2 FSR1.FPNR2.F a1 a2 a3 b1 b2 b3 c1 c2 c3 d1 d2 d1 d2 d1 d2 b1 b2 b1 b2 b1 b2 R3 ::= R1 R2 R3

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15 Renaming Renaming is useful to resolve possible name conflicts R2 ::= (R2( ), R1) is a list of terms of the form A O A N Returns R1 with columns renamed according to Simplified notation: R2 ::= R1 F (F) F F

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16 Example: Renaming IDNameMajor 083920 183280 Adam Saniya Math CS Student Students ::= (Name Fullname, Major Plan) Student IDFullnamePlan 083920 183280 Adam Saniya Math CS Students

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17 Joins The most commonly used way to combine several relations Condition Joins (Theta-join) Equijoin –Natural Join Outer join One of the most useful operators in R.A.

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18 Theta-Join Theta-join joins two relations and extracts some rows from the result R3 ::= R1 C R2 = C (R1 R2) Common special case: C consists solely of equalities (Equijoin) The schema of the result consists of attributes of R1 followed by those attributes of R2 that do not appear in C.

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19 Example: Theta-Join FoodserviceFoodPrice Bon Appetite Brubaker’s Bon Appetite Mofongo Meatball 4.50 5.00 5.50 CheapFood NameCuisine Sushi Meatball Mofongo Japanese Global Puerto Rican Food R1 ::= Food Price<5.50 Name=Food CheapFood NameCuisineFoodserviceFoodPrice Meatball Mofongo Global Puerto Rican Brubaker’s Bon Appetite Meatball Mofongo 5.00 4.50 R1

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20 Natural Join A further special case R3 ::= R1 R2 = A’ ( C’ (R1 R2) ) An Equijoin in which equalities are specified on all common attributes of R1 and R2 The result is guaranteed not to have attributes with the same name

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21 Example: Natural Join FoodserviceFoodPrice Bon Appetite Brubaker’s Bon Appetite Mofongo Meatball 4.50 5.00 5.50 CheapFood FoodCuisine Sushi Meatball Mofongo Japanese Global Puerto Rican Food1 R1 ::= Food1 CheapFood R1 FoodCuisineFoodservicePrice Meatball Mofongo Global Puerto Rican Brubaker’s Bon Appetite 5.00 5.50 4.50

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22 Division Division. R3 ::= R1 / R2 It is not needed as often But sometimes simplifies a query significantly “Find Foodservices that provide all foods” Properties: R2’s attributes R1’s attributes R3’s attributes = R1’s attributes – R2’s attributes R3 contains all tuples (T3) s.t. for every tuple T2 R2, T3.T2 T1 (set of tuples of R1) In other words, R3 R2 R1

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23 Examples: Division A1A2A3 1121211212 1213312133 1111111111 R1 A1A3 1212 1111 R2 A2 123123 R3 A1A3 1212 1111 R4 A2 1313 R5 A1A3 11 R6 R3 ::= R1 / R2 R4 ::= R1 / R3 R6 ::= R1 / R5 * * * * * * * * * * * *

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24 Complex Expressions Expressions with several operators Example: Precedence of operators: 1. Select, Project, Renaming 2. Products, Division, Joins 3. Intersection 4. Union, Difference Therefore, we could write: How about this? R2 ::= Foodservice ( price<6.0 (Provides Food1)) R2 ::= Foodservice price<6.0 (Provides Food1) R2 ::= Foodservice price<6.0 Provides Food1

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25 Expression Trees Example: “Find all Foodservice names that provide meatball for less that 6 bucks and their rate is over 6. Food=“Meatball” Food1 Price<6.0 Provides FoodService Rate>6 Name Useful when writing complex queries

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26 Hands-on Example 1 “Find all students who like Mofongo and frequent Pollo Tropical”. (Contest 1)

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27 Hands-on Example 2 “Find names of all ICOM students who do not frequent a foodservice that its rate is greater than 8” (Contest 2)

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28 Hands-on Example 3 “Find all Foodservice names that provide at least two foods with the same price” (Contest 3)

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29 Relational Algebra on Bags Bags are a more general data structure Some operations are more efficient on bags Following R.A. operators produce a different result on a bag than on a set Projection, Division, Union, Intersection, Difference Bag projection and bag division do not eliminate duplicates from the result. Recall: SQL tables are multisets (bags)

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30 U B, ∩ B, – B Bag Union. R3::= R1 U B R2 R3: all tuples of R1 and R2 Example: {1,1,2} U B {1,1,1,3}= {1,1,1,1,1,2,3} Bag Intersection. R3 ::= R1 ∩ B R2 R3: common tuples of R1 and R2 with the minimum show up Example: {1,1,2} ∩ B {1,1,1,3} = {1,1} Bag Difference. R3 ::= R1 – B R3 R3: all tuples appear in R1 – # of times appear in R2 Example: {1,1,2} – B {1,1,1,3} = {2} Do sets laws hold for bags?

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31 Extended R.A. Distinct eliminates duplicates from R1 R2 ::= (R1) Order sorts R1 based on list of attributes in A R2 ::= A (R1) Extended Projection R2 ::= A (R1) A consists of some attributes of R1, as well as (arithmetic) expressions and attribute renaming

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32 Examples: , , A1A2 12121212 21232123 R1 A1A2 122122 213213 R2 = (R1) A1A2A3 1121211212 1212312123 5545155451 R3 A1A2A3 1211212112 1122311223 5455154551 R4= A2,A1 (R3) R5 = A1 NewA1, A1, A1+A2,A2 (R1) NewA1A1A1+A2A2 12121212 12121212 33353335 21232123

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33 Extended R.A. cont’d Agg aggregates values of an attribute to compute a function Sum(A), Avg(A), Count(A), Min(A), Max(A) Not applicable on a relation, rather on an attribute Group groups tuples based on attributes R2 ::= A (R1) Where, A consists of some attributes of R1 may include some aggregate functions

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34 Examples: Agg, A1A2A3 1121211212 1232312323 5545155451 R1 R2 = A1,A2, Avg(A3), Sum(A2) (R1) Sum (A1) = 7 Count (A1) = 5 Avg (A3) = 4 A1A2Avg(A3)Sum(A2) 112112 123123 5 2.5 146146

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35 Extended R.A. cont’d Outer join R3 ::= R1 O R2 An extended natural join in which no tuple from R1 or R2 is omitted. Example: A1A2 122122 213213 R1 A2A3 234234 123123 R2 A1A2A3 122122 21342134 123123 R1 O R2

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36 Relational Completeness Examples Safe Relational Calculus R.A. SQL (has additional expressive power) Recommended Exercise: from Ramakrishnan’s Definition. A query language is at least as expressive as core relational algebra is said to be relationally complete. 4.1—4.7 (ignore Relational Calculus)

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