# CS 319: Theory of Databases: C4

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CS 319: Theory of Databases: C4
Dr. Alexandra I. Cristea

(provisionary) Content
Generalities DB Integrity constraints (FD revisited) Relational Algebra (revisited) Query optimisation Tuple calculus Domain calculus Query equivalence LLJ, DP and applications Temporal Data The Askew Wall

… previous FD

Relational model E.F. Codd: check wikipedia! Papers:
Reprint at: A relational Model of Data for Large Shared Data Banks Relational Completeness of Data Base Sublanguages

Relational Model* Structure of Relational Databases
Fundamental Relational-Algebra-Operations Additional Relational-Algebra-Operations *based on Silberschatz, Korth, Sudarshan Database System Concepts 5th Edition, Chapter 2

Example of a Relation Instance

Basic Structure Formally, given sets D1, D2, …. Dn a relation r is a subset of D1 x D2 x … x Dn Thus, a relation is a set of n-tuples (a1, a2, …, an) where each ai  Di

Example If customer_name = {Jones, Smith, Curry, Lindsay, …} /* Set of all customer names */ customer_street = {Main, North, Park, …} /* set of all street names*/ customer_city = {Harrison, Rye, Pittsfield, …} /* set of all city names */ Then r = { (Jones, Main, Harrison), (Smith, North, Rye), (Curry, North, Rye), (Lindsay, Park, Pittsfield) } is a relation over customer_name x customer_street x customer_city

Attribute Types Each attribute of a relation has a name
The set of allowed values for each attribute is called the domain of the attribute Attribute values are (normally) required to be atomic; that is, indivisible E.g. the value of an attribute can be an account number, but cannot be a set of account numbers Domain is said to be atomic if all its members are atomic The special value null is a member of every domain The null value causes complications in the definition of many operations We shall ignore the effect of null values in our main presentation and consider their effect later

Relation Schema A1, A2, …, An are attributes
R = (A1, A2, …, An ) is a relation schema Example: Customer_schema = (customer_name, customer_street, customer_city) r(R) denotes a relation r on the relation schema R customer (Customer_schema)

Relation Instance The current values (relation instance) of a relation are specified by a table An element t of r is a tuple, represented by a row in a table attributes (or columns) customer_name customer_street customer_city Jones Smith Curry Lindsay Main North Park Harrison Rye Pittsfield tuples (or rows) customer

Relations are Unordered
Order of tuples is irrelevant (tuples may be stored in an arbitrary order) Example: account relation with unordered tuples

Database A database consists of multiple relations
Information about an enterprise is broken up into parts, with each relation storing one part of the information Storing all information as a single relation such as bank(account_number, balance, customer_name, ..) results in repetition of information e.g.,if two customers own an account (What gets repeated?) the need for null values e.g., to represent a customer without an account Normalization theory deals with how to design relational schemas A database consists of multiple relations Information about an enterprise is broken up into parts, with each relation storing one part of the information account : stores information about accounts depositor : stores information about which customer owns which account customer : stores information about customers Storing all information as a single relation such as bank(account_number, balance, customer_name, ..) results in repetition of information e.g.,if two customers own an account (What gets repeated?) the need for null values e.g., to represent a customer without an account Normalization theory deals with how to design relational schemas

The customer Relation

The depositor Relation

Keys Let K  R K is a superkey of R if values for K are sufficient to identify a unique tuple of each possible relation r(R) Example: {customer_name, customer_street} and {customer_name} are both superkeys of Customer, if no two customers can possibly have the same name In real life, an attribute such as customer_id would be used instead of customer_name to uniquely identify customers, but we omit it to keep our examples small, and instead assume customer names are unique. Let K  R K is a superkey of R if values for K are sufficient to identify a unique tuple of each possible relation r(R) by “possible r ” we mean a relation r that could exist in the enterprise we are modeling. Example: {customer_name, customer_street} and {customer_name} are both superkeys of Customer, if no two customers can possibly have the same name

Keys (Cont.) K is a candidate key if K is minimal
Example: {customer_name} : superkey + no subset of it is a superkey. Primary key: a candidate key chosen as the principal means of identifying tuples within a relation K is a candidate key if K is minimal Example: {customer_name} is a candidate key for Customer, since it is a superkey and no subset of it is a superkey. Primary key: a candidate key chosen as the principal means of identifying tuples within a relation Should choose an attribute whose value never, or very rarely, changes. E.g. address is unique, but may change

Foreign Keys A relation schema may have an attribute that corresponds to the primary key of another relation. The attribute is called a foreign key. A relation schema may have an attribute that corresponds to the primary key of another relation. The attribute is called a foreign key. E.g. customer_name and account_number attributes of depositor are foreign keys to customer and account respectively. Only values occurring in the primary key attribute of the referenced relation may occur in the foreign key attribute of the referencing relation. Schema diagram

Query Languages Language in which user requests information from the database. Categories of languages Procedural Non-procedural, or declarative “Pure” languages: Relational algebra Tuple relational calculus Domain relational calculus Pure languages form underlying basis of query languages that people use.

Relational Algebra Procedural language Six basic operators
select:  project:  union:  set difference: – Cartesian product: x rename:  The operators take one or two relations as inputs and produce a new relation as a result.

Select Operation – Example
Relation r A B C D 1 5 12 23 7 3 10 A=B ^ D > 5 (r) A B C D 1 23 7 10

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: <attribute> op <attribute> or <constant> where op is one of: =, , >, . <.  Example of selection:  branch_name=“Perryridge”(account)

Project Operation – Example
Relation r: A B C 10 20 30 40 1 2 A,C (r) A C A C 1 2 1 2 =

Project Operation Notation: where A1, A2 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 Example: To eliminate the branch_name attribute of account account_number, balance (account)

Union Operation – Example
Relations r, s: A B A B 1 2 2 3 s r A B 1 2 3 r  s:

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 (example: 2nd column of r deals with the same type of values as does the 2nd column of s) Example: to find all customers with either an account or a loan customer_name (depositor)  customer_name (borrower)

Set Difference Operation – Example
Relations r, s: A B A B 1 2 2 3 s r r – s: A B 1

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

Cartesian-Product Operation – Example
B C D E Relations r, s: 1 2 10 20 a b r s A B C D E r x s: 1 2 10 20 a b

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.

Composition of Operations
Can build expressions using multiple operations Example: A=C(r x s) r x s A=C(r x s) A B C D E 1 2   10 20 a b A B C D E 1 2 10 20 a b

Rename Operation Allows us to refer to results of RA expressions & to refer to a relation by more than one name. Example:  x (E) returns the expression E under the name X expression E under name X, with attributes renamed to A1 , A2 , …., An . 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 returns the result of expression E under the name X, and with the attributes renamed to A1 , A2 , …., An .

Revise basic operators!

Relational Algebra Operators thus …
Procedural language with six basic operators select:  project:  union:  set difference: – Cartesian product: x rename:  The operators take one or two relations as inputs and produce a new relation as a result.

Banking Example branch (branch_name, branch_city, assets)
customer (customer_name, customer_street, customer_city) account (account_number, branch_name, balance) loan (loan_number, branch_name, amount) depositor (customer_name, account_number) borrower (customer_name, loan_number)

Example Queries amount > 1200 (loan)
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)) Find the names of all customers who have a loan, an account, or both, from the bank customer_name (borrower)  customer_name (depositor)

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) How can we simplify the second query? (hint: we don’t need the loan table – see previous example for loan customers on the previous page)

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)) There is not always a unique answer to a query. Here we see two possible example answers for the same query.

Example Queries Find the largest account balance Strategy:
Find those balances that are not the largest Rename account relation as d so that we can compare each account balance with all others Use set difference to find those account balances that were not found in the earlier step. The query is: balance(account) - account.balance (account.balance < d.balance (account x rd (account)))

Formal Definition A basic expression in the relational algebra consists of either one of the following: A relation in the database A constant relation Let E1 and E2 be relational-algebra expressions; the following are all relational-algebra expressions: E1  E2 E1 – E2 E1 x E2 p (E1), P is a predicate on attributes in E1 s(E1), S is a list consisting of some of the attributes in E1  x (E1), x is the new name for the result of E1 The operators therefore are defined on the relational algebra expressions and the result also maps onto the relational algebra expressions.

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

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) Please also note: r  s = s – (s – r) (due to symmetry of the intersection) So we could write the query with the basic operators – it only would be quite complex.

Set-Intersection Operation – Example
Relation r, s: r  s A B A B 1 2 2 3 r s A B

Natural-Join Operation
Notation: r s Let r and s be relations on schemas R and S respectively. Then, r s is a relation on schema R  S obtained as follows: Consider each pair of tuples tr from r and ts from s. If tr and ts have the same value on each of the attributes in R  S, add a tuple t to the result, where t has the same value as tr on r t has the same value as ts on s

Natural joint example 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))

Natural Join Operation – Example
Relations r, s: B D E A B C D 1 3 2 a b 1 2 4 a b r s A B C D E r s 1 2 a b

Division Operation r  s Notation:
Suited to queries that include the phrase “for all”*. Let r and s be relations on schemas R and S respectively where R = (A1, …, Am , B1, …, Bn ) S = (B1, …, Bn) The result of r  s is a relation on schema R – S = (A1, …, Am) r  s = { t | t   R-S (r)   u  s ( tu  r ) } Where tu means the concatenation of tuples t and u to produce a single tuple *Please note that whilst the division operator suits some queries that include ‘for all’, it doesn’t work for all queries that include ‘for all’!! (only for the ones where the set of values is constant!)

Division Operation – Example
B B 1 2 3 4 6 Relations r, s: 1 2 s A r  s: r

Another Division Example
Relations r, s: A B C D E D E a a b 1 3 a b 1 s r r  s: A B C a

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. If we want to express the ‘for all tuples something holds’ expression with the basic operators, it is easier to think of the negation: “there is at least one tuple for which something doesn’t hold”. So, if “for all tuples in s(S) we have a corresponding tuple in r(R)”, we rewrite this as: The negation of “there is at least one tuple in s(S) for which we don’t have a corresponding tuple in r(R)” = The negation of “there is a tuple in s(S) for which we don’t have a corresponding tuple in r(R)” = The negation of AuxilliaryQuery = = Tuples in (a restriction of) r(R) minus AuxilliaryQuery = = Tuples in (a restriction of) r(R) - AuxilliaryQuery = Now we have to compute the AuxilliaryQuery = “there is a tuple in s(S) for which we don’t have a corresponding tuple in r(R)” = = (extension of) s(S) minus tuples in r(R) = = (extension of) s(S) - tuples in r(R)

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.

Bank Example Queries Find the names of all customers who have a loan and an account at the bank. customer_name (borrower)  customer_name (depositor) Find the name of all customers who have a loan at the bank and the loan amount customer_name, loan_number, amount (borrower loan) Please look at the first query and compare with the previous query that looked at a loan, account, or both. What is different? Why do we use reunion in one query, and intersection in the other one? (Answer: here we want customers who have both a loan and an account. We don’t count the customers who have only an account, for instance.)

Bank Example Queries Find all customers who have an account from at least the “Downtown” and the “Uptown” branches. Query 1 customer_name (branch_name = “Downtown” (depositor account ))  customer_name (branch_name = “Uptown” (depositor account)) Query 2 customer_name, branch_name (depositor account)  temp(branch_name) ({(“Downtown” ), (“Uptown” )}) Note that Query 2 uses a constant relation.

Bank Example Queries 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)) Here we can divide by the branch name table, because it is a constant.

Library Case reservation book author name department Initials title
publisher date year cancelled copy borrow name PK = primary key FK = foreign key department department from cpYear to present

Library database book ( ISBN, title, publisher, year )
author (ISBN, initials, name ) copy (barcode, ISBN, department, cpYear, present ) reservation (name, department, ISBN, date, cancelled ) borrow (name, department, barcode, from, to )

Library Questions (RA)
List all the authors of books of which the library has a copy that has never been borrowed. List all the authors of books that have never been borrowed. List all the authors of which no book has ever been borrowed. In these questions we need to find “borrow” records that correspond to books of an author. The “borrow” records only contain the barcode, not the ISBN. So the “copy” table is needed to make the link between the book and borrowed copies of the book. Note the difference between the three questions. In the first we look at just one copy. In the second we look at all the copies of a book. In the third we look at all the books written by an author. Each time we need the “difference” operator, but on a different subquery.

(simple) Employee database
employee(person_name, street, city) works(person_name, company_name, salary) company(company_name, city) manages(person_name, manager_name)

Exercise 2.1.c Find the names of all employees who earn more than every employee of “SBC”. The “more than every...” clause cannot be expressed directly in the relational algebra. For all other employees there is an employee of “SBC” earning more. This can be described using a selection on a cartesian product (of works with works). We then use the difference to find the correct result. The query we really answer is: “List the names of all the employees, except those for which there is an employee of “SBC” with a higher or equal salary.” This is Exercise 2.1.c from the Silberschatz book, fifth Edition.

Solution 2.1.c person-name (W) –
- person-name( (w.salary<=w2.salary ^ w2.company-name = ‘SBC’ (W x W2(W) ))

Exercise 2.5.e Find all companies located in every city in which “SBC” is located.

Case 1: 1 city per company C.company-name ( (C.city=SBC.city
(C (company ) x (SBC.company-name=‘SBC’ (SBC (company ) ) ) Trivial!

Case 2: multiple cities per company
Find all companies located in every city in which “SBC” is located. The “every city...” clause cannot be expressed directly in the relational algebra (except for division). Let’s see …

Case 2a: multiple cities: division
All cities where SBC is located: SBCCity = city (company-name=‘SBC’ (company ) ) Is SBCCity constant? Yes! So we can use it on the right hand side of the division. companies located in every city in which “SBC” is located: company  SBCCity Still quite neat!!

Case 2b: multiple cities: no division
Find all companies located in every city in which “SBC” is located. The “every city...” clause cannot be expressed directly in the relational algebra (except for division). For the other companies there is a city in which “SBC” is located and the other company is not. The cities where a company is not located are all the cities except the ones where the company is located. We thus need 2 times a difference in this query. Can also be formulated using a difference operator This is really a question on the division operator. Doing it without it forces us to really understand the question well. We really would like to answer the question: “List all the companies, except those for which there is a city in which “SBC” is located and the other company is not”, but the difficulty then is in ensuring that the first and last company are the same. We can perform a trick here: First lets do the following: “List every company with every city where SBC is located.” So we replace the cities where the companies are located with the cities where SBC is located. Now “Subtract from this the list of companies with the cities where they are located.” This yields a list of companies with every city in which SBC is located and they are not. Now we subtract these companies from the whole list of companies and then we have the companies for which there is no city in which SBC is located and they are not.

Case 2b: multiple cities: no division
For the other companies there is a city in which “SBC” is located and the other company is not = IntRes. E = company-name (company ) x SBCCity IntRes = E – company We want not (IntRes): company-name (company ) - company-name (IntRes)

What is the meaning of: SBCCity 

Recognizing Types of Queries
Identify the type of the following queries, and afterwards also translate them to the algebra (PSJ, union, intersection, difference, division): Give the name of customers that have a loan with a branch where they also have an account. Give the name of customers who have a loan at a branch where they do not have an account. Give the name of customers who have a loan at every branch where they have an account. Give the name of customers who have loans only at branches where they have an account. In this course we are going to try to understand to recognize different types of queries by looking at the question formulated in natural language. We only establish the type. Writing out the queries in RA is homework. First: PSJ: we search for a match between a customer and a loan and an account. As soon as one match is found we have a tuple for the result. We need to recognize PSJ queries right away by checking that the query works by matching one tuple at a time from each relation that is used. Second: difference: It is important to realize here what the two subqueries (before and after the difference operator) should be. The word “not” in the query means that a difference operator is needed. Third. complex difference query. Whenever we see “every” we should consider whether it is a division query or not. Division means that the “every” should refer to a constant set of tuples. In this query the set depends on the custommer (or where he has an account) so it is not a division query. A standard approach to such a question is to rephrase the question to get reach of the “every” (or “each” if that word is used). Give the names of customers for whom there is no branch where they have an account but no loan. (This is the complement of question 2 with loan and account reversed.) Fourth: This is again a difference question. These are all the customers except those who have a loan at a branch where they do not have an account. It is thus the complement of question 2.

account  depositor  borrower  loan)) 6. branch-name(branch)  branch-name(branch-city  customer-city( branch  account  depositor  customer)) 7. X.customer-name (X.account-number = Y.account-number  X.customer-name  Y.customer-name (X(depositor)  Y(depositor))) First: Give the name of the customers who have an account with a higher balance than the amount of one of the loans of these customers at the same branch. Second: Give the name of the branches that only have depositors from their own city. Third: Give the name of customers who have a joint account with someone else.

Beer Database visits(drinker, bar) serves(bar, beer)
likes(drinker, beer).

Beer questions with a difference
Give all drinkers that visit bars that don’t serve any beer they like Give all drinkers that only visit bars that serve a beer they like Give all drinkers that only visit bars that serve no beer they like Give all drinkers that only visit bars that serve all beers they like (and maybe other beers as well) Give all drinkers that only visit bars that only serve beers they like (and thus serve nothing else) These questions are related but differ by the use and position of the ‘not’ and ‘only’ keywords. Give all drinkers that visit bars that don’t serve a beer they like: This is a question about information that is not directly available in the database. A drinker qualifies when he visits one bar where no delectable beer is served. So we take the Visits relation and subtract the combinations of bars and drinkers that both favour the same beer: Proj_(d)(Visits – Proj_(d, b)(Visits join Serves join Likes)) Give all drinkers that only visit bars that serve a beer they like The issue here is that the drinker never visits a bar that doesn’t serve a beer he likes. So we have to exclude the drinkers emerging from the previous question from the set of all drinkers: Proj_d(Visits) - Proj_(d)(Visits – Proj_(d, b)(Visits join Serves join Likes)) Give all drinkers that only visit bars that serve no beer they like In this case we again are looking at the difference of the set of all drinkers and the one with the drinkers that visit bars that serve a beer they like: Proj_d(Visits) – Proj_d(Visits join Servers join Likes) Give all drinkers that only visit bars that only serve beers they like This question involves twice the adjective ‘only’. The first only means that we consider the difference between the set of all drinkers and the set of drinkers that not only visit bars that only serve beers they like. That is the set of all drinkers minus the set of drinkers that also visit bars that also serve beers they don’t like: Proj_d(Visits) – Proj_d(Visits join Serves join (Proj_d(Likes) times Proj_b(Likes) - Likes))

Summary We have learned RA
We have learned to perform simple and more complex queries in RA