Relational Algebra and SQL Prof. Sin-Min Lee Department of Computer Science San Jose State University.

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Relational Algebra and SQL Prof. Sin-Min Lee Department of Computer Science San Jose State University

©Silberschatz, Korth and Sudarshan3.2Database System Concepts Functional Dependencies A form of constraint  hence, part of the schema Finding them is part of the database design Also used in normalizing the relations Warning: this is the most abstract, and “hardest” part of the course.

©Silberschatz, Korth and Sudarshan3.3Database System Concepts Functional Dependencies Definition: If two tuples agree on the attributes then they must also agree on the attributes Formally: A 1, A 2, …, A n  B 1, B 2, …, B m A 1, A 2, …, A n B 1, B 2, …, B m

©Silberschatz, Korth and Sudarshan3.4Database System Concepts Examples EmpID  Name, Phone, Position Position  Phone but Phone  Position EmpIDNamePhonePosition E0045Smith1234Clerk E1847John9876Salesrep E1111Smith9876Salesrep E9999Mary1234Lawyer

©Silberschatz, Korth and Sudarshan3.5Database System Concepts In General To check A  B, erase all other columns check if the remaining relation is many-one (called functional in mathematics)

©Silberschatz, Korth and Sudarshan3.6Database System Concepts Example EmpIDNamePhonePosition E0045Smith1234Clerk E1847John9876Salesrep E1111Smith9876Salesrep E9999Mary1234Lawyer Position  Phone

©Silberschatz, Korth and Sudarshan3.7Database System Concepts Typical Examples of FDs Product: name  price, manufacturer Person: ssn  name, age Company: name  stockprice, president

©Silberschatz, Korth and Sudarshan3.8Database System Concepts Example Product(name, category, color, department, price) name  color category  department color, category  price name  color category  department color, category  price Consider these FDs: What do they say ?

©Silberschatz, Korth and Sudarshan3.9Database System Concepts Example FD’s are constraints on relations: On some instances they hold On others they don’t namecategorycolordepartmentprice GizmoGadgetGreenToys49 TweakerGadgetGreenToys99 Does this instance satisfy all the FDs ? name  color category  department color, category  price name  color category  department color, category  price

©Silberschatz, Korth and Sudarshan3.10Database System Concepts Example namecategorycolordepartmentprice GizmoGadgetGreenToys49 TweakerGadgetBlackToys99 GizmoStationaryGreenOffice-supp.59 What about this one ? name  color category  department color, category  price name  color category  department color, category  price

©Silberschatz, Korth and Sudarshan3.11Database System Concepts Example If some FDs are satisfied, then others are satisfied too If all these FDs are true: name  color category  department color, category  price name  color category  department color, category  price Then this FD also holds: name, category  price Why ??

©Silberschatz, Korth and Sudarshan3.12Database System Concepts Inference Rules for FD’s Is equivalent to Splitting rule and Combining rule A1...AmB1...Bm A 1, A 2, …, A n  B 1, B 2, …, B m A 1, A 2, …, A n  B 1 A 1, A 2, …, A n  B A 1, A 2, …, A n  B m A 1, A 2, …, A n  B 1 A 1, A 2, …, A n  B A 1, A 2, …, A n  B m

©Silberschatz, Korth and Sudarshan3.13Database System Concepts Inference Rules for FD’s (continued) Trivial Rule Why ? A1A1 …AmAm where i = 1, 2,..., n A 1, A 2, …, A n  A i

©Silberschatz, Korth and Sudarshan3.14Database System Concepts Inference Rules for FD’s (continued) Transitive Closure Rule If and then Why ? A 1, A 2, …, A n  B 1, B 2, …, B m B 1, B 2, …, B m  C 1, C 2, …, C p A 1, A 2, …, A n  C 1, C 2, …, C p

©Silberschatz, Korth and Sudarshan3.15Database System Concepts A1A1 …AmAm B1B1 …BmBm C1C1...CpCp

©Silberschatz, Korth and Sudarshan3.16Database System Concepts Example (continued) Start from the following FDs: Infer the following FDs: 1. name  color 2. category  department 3. color, category  price 1. name  color 2. category  department 3. color, category  price Inferred FD Which Rule did we apply ? 4. name, category  name 5. name, category  color 6. name, category  category 7. name, category  color, category 8. name, category  price

©Silberschatz, Korth and Sudarshan3.17Database System Concepts Example (continued) Answers: Inferred FD Which Rule did we apply ? 4. name, category  nameTrivial rule 5. name, category  colorTransitivity on 4, 1 6. name, category  categoryTrivial rule 7. name, category  color, categorySplit/combine on 5, 6 8. name, category  priceTransitivity on 3, 7 1. name  color 2. category  department 3. color, category  price 1. name  color 2. category  department 3. color, category  price

©Silberschatz, Korth and Sudarshan3.18Database System Concepts Another Example Enrollment(student, major, course, room, time) student  major major, course  room course  time What else can we infer ?

©Silberschatz, Korth and Sudarshan3.19Database System Concepts Another Rule If then Augmentation follows from trivial rules and transitivity How ? A 1, A 2, …, A n  B A 1, A 2, …, A n, C 1, C 2, …, C p  B Augmentation

©Silberschatz, Korth and Sudarshan3.20Database System Concepts R(A B C D) Functional Dependency Graph A B C D

©Silberschatz, Korth and Sudarshan3.21Database System Concepts Problem: infer ALL FDs Given a set of FDs, infer all possible FDs How to proceed ? Try all possible FDs, apply all 3 rules  E.g. R(A, B, C, D): how many FDs are possible ? Drop trivial FDs, drop augmented FDs  Still way too many Better: use the Closure Algorithm (next)

©Silberschatz, Korth and Sudarshan3.22Database System Concepts Relational Algebra Procedural language Six basic operators  select  project  union  set difference  Cartesian product  rename The operators take two or more relations as inputs and give a new relation as a result.

©Silberschatz, Korth and Sudarshan3.23Database System Concepts Select Operation – Example Relation r ABCD    A=B ^ D > 5 (r) ABCD  

©Silberschatz, Korth and Sudarshan3.24Database System Concepts Select Operation Notation:  p (r) p is called the selection predicate Defined as:  p ( r) = {t | t  r and p(t)} Where p is a formula in propositional calculus consisting of terms connected by :  (and),  (or),  (not) Each term is one of: op or where op is one of: =, , >, . <.  Example of selection:  branch-name=“Perryridge ” (account)

©Silberschatz, Korth and Sudarshan3.25Database System Concepts Project Operation – Example Relation r: ABC  AC  = AC   A,C (r)

©Silberschatz, Korth and Sudarshan3.26Database System Concepts Project Operation Notation:  A1, A2, …, Ak (r) where A 1, A 2 are attribute names and r is a relation name. The result is defined as the relation of k columns obtained by erasing the columns that are not listed Duplicate rows removed from result, since relations are sets E.g. To eliminate the branch-name attribute of account  account-number, balance (account)

©Silberschatz, Korth and Sudarshan3.27Database System Concepts Union Operation – Example Relations r, s: r  s: AB  AB  2323 r s AB 

©Silberschatz, Korth and Sudarshan3.28Database System Concepts Union Operation Notation: r  s Defined as: r  s = {t | t  r or t  s} For r  s to be valid. 1. r, s must have the same arity (same number of attributes) 2. The attribute domains must be compatible (e.g., 2nd column of r deals with the same type of values as does the 2nd column of s) E.g. to find all customers with either an account or a loan  customer-name (depositor)   customer-name (borrower)

©Silberschatz, Korth and Sudarshan3.29Database System Concepts Set Difference Operation – Example Relations r, s: r – s : AB  AB  2323 r s AB  1111

©Silberschatz, Korth and Sudarshan3.30Database System Concepts Set Difference Operation Notation r – s Defined as: r – s = {t | t  r and t  s} Set differences must be taken between compatible relations.  r and s must have the same arity  attribute domains of r and s must be compatible

©Silberschatz, Korth and Sudarshan3.31Database System Concepts Cartesian-Product Operation-Example Relations r, s: r x s: AB  1212 AB  CD  E aabbaabbaabbaabb CD  E aabbaabb r s

©Silberschatz, Korth and Sudarshan3.32Database System Concepts Cartesian-Product Operation Notation r x s Defined as: r x s = {t q | t  r and q  s} Assume that attributes of r(R) and s(S) are disjoint. (That is, R  S =  ). If attributes of r(R) and s(S) are not disjoint, then renaming must be used.

©Silberschatz, Korth and Sudarshan3.33Database System Concepts Composition of Operations Can build expressions using multiple operations Example:  A=C (r x s) r x s  A=C (r x s) AB  CD  E aabbaabbaabbaabb ABCDE   20 aabaab

©Silberschatz, Korth and Sudarshan3.34Database System Concepts Rename Operation Allows us to name, and therefore to refer to, the results of relational-algebra expressions. Allows us to refer to a relation by more than one name. Example:  x (E) returns the expression E under the name X If a relational-algebra expression E has arity n, then  x (A1, A2, …, An) (E) returns the result of expression E under the name X, and with the attributes renamed to A 1, A2, …., An.

©Silberschatz, Korth and Sudarshan3.35Database System Concepts Banking Example branch (branch-name, branch-city, assets) customer (customer-name, customer-street, customer-only) account (account-number, branch-name, balance) loan (loan-number, branch-name, amount) depositor (customer-name, account-number) borrower (customer-name, loan-number)

©Silberschatz, Korth and Sudarshan3.36Database System Concepts Example Queries Find all loans of over $1200  amount > 1200 (loan) Find the loan number for each loan of an amount greater than $1200  loan-number (  amount > 1200 (loan))

©Silberschatz, Korth and Sudarshan3.37Database System Concepts Example Queries Find the names of all customers who have a loan, an account, or both, from the bank  customer-name (borrower)   customer-name (depositor) Find the names of all customers who have a loan and an account at bank.  customer-name (borrower)   customer-name (depositor)

©Silberschatz, Korth and Sudarshan3.38Database System Concepts Example Queries Find the names of all customers who have a loan at the Perryridge branch.  customer-name (  branch-name=“Perryridge ” (  borrower.loan-number = loan.loan-number (borrower x loan))) Find the names of all customers who have a loan at the Perryridge branch but do not have an account at any branch of the bank.  customer-name (  branch-name = “Perryridge” (  borrower.loan-number = loan.loan-number ( borrower x loan ))) –  customer-name ( depositor )

©Silberschatz, Korth and Sudarshan3.39Database System Concepts Example Queries Find the names of all customers who have a loan at the Perryridge branch.  Query 1  customer-name (  branch-name = “Perryridge” (  borrower.loan-number = loan.loan-number (borrower x loan)))  Query 2  customer-name (  loan.loan-number = borrower.loan-number ( (  branch-name = “Perryridge” (loan)) x borrower) )

©Silberschatz, Korth and Sudarshan3.40Database System Concepts Example Queries Find the largest account balance Rename account relation as d The query is:  balance (account) -  account.balance (  account.balance < d.balance (account x  d (account)))

©Silberschatz, Korth and Sudarshan3.41Database System Concepts Additional Operations We define additional operations that do not add any power to the relational algebra, but that simplify common queries. Set intersection Natural join Division Assignment

©Silberschatz, Korth and Sudarshan3.42Database System Concepts Set-Intersection Operation Notation: r  s Defined as: r  s ={ t | t  r and t  s } Assume:  r, s have the same arity  attributes of r and s are compatible Note: r  s = r - (r - s)

©Silberschatz, Korth and Sudarshan3.43Database System Concepts Set-Intersection Operation - Example Relation r, s: r  s A B   2323 r s  2

©Silberschatz, Korth and Sudarshan3.44Database System Concepts Natural-Join Operation Notation: r s Let r and s be relations on schemas R and S respectively.The result is a relation on schema R  S which is obtained by considering 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 R  S, a tuple t is added to the result, where  t has the same value as t r on r  t has the same value as t s on s Example: R = (A, B, C, D) S = (E, B, D) Result schema = (A, B, C, D, E) r s is defined as:  r.A, r.B, r.C, r.D, s.E (  r.B = s.B r.D = s.D (r x s))

©Silberschatz, Korth and Sudarshan3.45Database System Concepts Natural Join Operation – Example Relations r, s: AB  CD  aababaabab B D aaabbaaabb E  r AB  CD  aaaabaaaab E  s r s

©Silberschatz, Korth and Sudarshan3.46Database System Concepts Division Operation Suited to queries that include the phrase “for all”. Let r and s be relations on schemas R and S respectively where  R = (A 1, …, A m, B 1, …, B n )  S = (B 1, …, B n ) The result of r  s is a relation on schema R – S = (A 1, …, A m ) r  s = { t | t   R-S (r)   u  s ( tu  r ) } r  s

©Silberschatz, Korth and Sudarshan3.47Database System Concepts Division Operation – Example Relations r, s: r  s:r  s: A B  1212 AB  r s

©Silberschatz, Korth and Sudarshan3.48Database System Concepts Another Division Example AB  aaaaaaaaaaaaaaaa CD  aabababbaabababb E Relations r, s: r  s:r  s: D abab E 1111 AB  aaaa C  r s

©Silberschatz, Korth and Sudarshan3.49Database System Concepts Division Operation (Cont.) Property  Let q – r  s  Then q is the largest relation satisfying q x s  r Definition in terms of the basic algebra operation Let r(R) and s(S) be relations, and let S  R r  s =  R-S (r) –  R-S ( (  R-S (r) x s) –  R-S,S (r)) To see why   R-S,S (r) simply reorders attributes of r   R-S (  R-S (r) x s) –  R-S,S (r)) gives those tuples t in  R-S (r) such that for some tuple u  s, tu  r.

©Silberschatz, Korth and Sudarshan3.50Database System Concepts Assignment Operation The assignment operation (  ) provides a convenient way to express complex queries, write query as a sequential program consisting of a series of assignments followed by an expression whose value is displayed as a result of the query. Assignment must always be made to a temporary relation variable. Example: Write r  s as temp1   R-S (r) temp2   R-S ((temp1 x s) –  R-S,S (r)) result = temp1 – temp2  The result to the right of the  is assigned to the relation variable on the left of the .  May use variable in subsequent expressions.

©Silberschatz, Korth and Sudarshan3.51Database System Concepts Example Queries Find all customers who have an account from at least the “Downtown” and the Uptown” branches.  Query 1  CN (  BN=“Downtown ” (depositor account))   CN (  BN=“Uptown ” (depositor account)) where CN denotes customer-name and BN denotes branch-name.  Query 2  customer-name, branch-name (depositor account)   temp(branch-name ) ({(“Downtown”), (“Uptown”)})

©Silberschatz, Korth and Sudarshan3.52Database System Concepts Find all customers who have an account at all branches located in Brooklyn city.  customer-name, branch-name (depositor account)   branch-name (  branch-city = “Brooklyn” (branch)) Example Queries

©Silberschatz, Korth and Sudarshan3.53Database System Concepts Extended Relational-Algebra-Operations Generalized Projection Outer Join Aggregate Functions

©Silberschatz, Korth and Sudarshan3.54Database System Concepts Generalized Projection Extends the projection operation by allowing arithmetic functions to be used in the projection list.  F1, F2, …, Fn (E) E is any relational-algebra expression Each of F 1, F 2, …, F n are are arithmetic expressions involving constants and attributes in the schema of E. Given relation credit-info(customer-name, limit, credit-balance), find how much more each person can spend:  customer-name, limit – credit-balance (credit-info)

©Silberschatz, Korth and Sudarshan3.55Database System Concepts Aggregate Functions and Operations Aggregation function takes a collection of values and returns a single value as a result. avg: average value min: minimum value max: maximum value sum: sum of values count: number of values Aggregate operation in relational algebra G 1, G 2, …, G n g F 1 ( A 1), F 2 ( A 2),…, F n ( A n) (E)  E is any relational-algebra expression  G 1, G 2 …, G n is a list of attributes on which to group (can be empty)  Each F i is an aggregate function  Each A i is an attribute name

©Silberschatz, Korth and Sudarshan3.56Database System Concepts Aggregate Operation – Example Relation r: AB   C g sum(c) (r) sum-C 27

©Silberschatz, Korth and Sudarshan3.57Database System Concepts Aggregate Operation – Example Relation account grouped by branch-name: branch-name g sum(balance) (account) branch-nameaccount-numberbalance Perryridge Brighton Redwood A-102 A-201 A-217 A-215 A branch-namebalance Perryridge Brighton Redwood

©Silberschatz, Korth and Sudarshan3.58Database System Concepts Aggregate Functions (Cont.) Result of aggregation does not have a name  Can use rename operation to give it a name  For convenience, we permit renaming as part of aggregate operation branch-name g sum(balance) as sum-balance (account)

©Silberschatz, Korth and Sudarshan3.59Database System Concepts Outer Join An extension of the join operation that avoids loss of information. Computes the join and then adds tuples form one relation that does not match tuples in the other relation to the result of the join. Uses null values:  null signifies that the value is unknown or does not exist  All comparisons involving null are (roughly speaking) false by definition.  Will study precise meaning of comparisons with nulls later

©Silberschatz, Korth and Sudarshan3.60Database System Concepts Outer Join – Example Relation loan loan-numberamount L-170 L-230 L Relation borrower customer-nameloan-number Jones Smith Hayes L-170 L-230 L-155 branch-name Downtown Redwood Perryridge

©Silberschatz, Korth and Sudarshan3.61Database System Concepts Outer Join – Example Inner Join loan Borrower loan borrower Left Outer Join loan-numberamount L-170 L customer-name Jones Smith branch-name Downtown Redwood loan-numberamount L-170 L-230 L customer-name Jones Smith null branch-name Downtown Redwood Perryridge

©Silberschatz, Korth and Sudarshan3.62Database System Concepts Outer Join – Example Right Outer Join loan borrower loan-numberamount L-170 L-230 L null customer-name Jones Smith Hayes loan-numberamount L-170 L-230 L-260 L null customer-name Jones Smith null Hayes loan borrower Full Outer Join branch-name Downtown Redwood null branch-name Downtown Redwood Perryridge null

©Silberschatz, Korth and Sudarshan3.63Database System Concepts Null Values It is possible for tuples to have a null value, denoted by null, for some of their attributes null signifies an unknown value or that a value does not exist. The result of any arithmetic expression involving null is null. Aggregate functions simply ignore null values  Is an arbitrary decision. Could have returned null as result instead.  We follow the semantics of SQL in its handling of null values For duplicate elimination and grouping, null is treated like any other value, and two nulls are assumed to be the same  Alternative: assume each null is different from each other  Both are arbitrary decisions, so we simply follow SQL

©Silberschatz, Korth and Sudarshan3.64Database System Concepts Null Values Comparisons with null values return the special truth value unknown  If false was used instead of unknown, then not (A = 5 Three-valued logic using the truth value unknown:  OR: (unknown or true) = true, (unknown or false) = unknown (unknown or unknown) = unknown  AND: (true and unknown) = unknown, (false and unknown) = false, (unknown and unknown) = unknown  NOT: (not unknown) = unknown  In SQL “P is unknown” evaluates to true if predicate P evaluates to unknown Result of select predicate is treated as false if it evaluates to unknown

©Silberschatz, Korth and Sudarshan3.65Database System Concepts Modification of the Database The content of the database may be modified using the following operations:  Deletion  Insertion  Updating All these operations are expressed using the assignment operator.

©Silberschatz, Korth and Sudarshan3.66Database System Concepts Deletion A delete request is expressed similarly to a query, except instead of displaying tuples to the user, the selected tuples are removed from the database. Can delete only whole tuples; cannot delete values on only particular attributes A deletion is expressed in relational algebra by: r  r – E where r is a relation and E is a relational algebra query.

©Silberschatz, Korth and Sudarshan3.67Database System Concepts Deletion Examples Delete all account records in the Perryridge branch. account  account –  branch-name = “Perryridge” (account) Delete all loan records with amount in the range of 0 to 50 loan  loan –   amount  0  and amount  50 (loan) Delete all accounts at branches located in Needham. r 1    branch-city = “Needham” (account branch) r 2   branch-name, account-number, balance (r 1 ) r 3   customer-name, account-number (r 2 depositor) account  account – r 2 depositor  depositor – r 3

©Silberschatz, Korth and Sudarshan3.68Database System Concepts Insertion To insert data into a relation, we either:  specify a tuple to be inserted  write a query whose result is a set of tuples to be inserted in relational algebra, an insertion is expressed by: r  r  E where r is a relation and E is a relational algebra expression. The insertion of a single tuple is expressed by letting E be a constant relation containing one tuple.

©Silberschatz, Korth and Sudarshan3.69Database System Concepts Insertion Examples Insert information in the database specifying that Smith has $1200 in account A-973 at the Perryridge branch. account  account  {(“Perryridge”, A-973, 1200)} depositor  depositor  {(“Smith”, A-973)} Provide as a gift for all loan customers in the Perryridge branch, a $200 savings account. Let the loan number serve as the account number for the new savings account. r 1  (  branch-name = “Perryridge” (borrower loan)) account  account   branch-name, account-number,200 (r 1 ) depositor  depositor   customer-name, loan-number,(r 1 )

©Silberschatz, Korth and Sudarshan3.70Database System Concepts Updating A mechanism to change a value in a tuple without charging all values in the tuple Use the generalized projection operator to do this task r   F1, F2, …, FI, (r) Each F, is either the ith attribute of r, if the ith attribute is not updated, or, if the attribute is to be updated F i is an expression, involving only constants and the attributes of r, which gives the new value for the attribute

©Silberschatz, Korth and Sudarshan3.71Database System Concepts Update Examples Make interest payments by increasing all balances by 5 percent. account   AN, BN, BAL * 1.05 (account) where AN, BN and BAL stand for account-number, branch-name and balance, respectively. Pay all accounts with balances over $10,000 6 percent interest and pay all others 5 percent account   AN, BN, BAL * 1.06 (  BAL  (account))   AN, BN, BAL * 1.05 (  BAL  (account))