3.1Database System Concepts Chapter 3: Relational Model Structure of Relational Databases Relational Algebra Tuple Relational Calculus Domain Relational.

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3.1Database System Concepts Chapter 3: Relational Model Structure of Relational Databases Relational Algebra Tuple Relational Calculus Domain Relational Calculus Extended Relational-Algebra-Operations

3.2Database System Concepts Example of a Relation

3.3Database System Concepts Basic Structure Formally, given sets D 1, D 2, …. D n a relation r is a subset of D 1 x D 2 x … x D n Thus a relation is a set of n-tuples (a 1, a 2, …, a n ) where each a i  D i Example: if customer-name = {Jones, Smith, Curry, Lindsay} customer-street = {Main, North, Park} customer-city = {Harrison, Rye, Pittsfield} 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

3.4Database System Concepts 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. multivalued attribute values are not atomic  E.g. composite attribute values are not 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

3.5Database System Concepts Relation Schema A 1, A 2, …, A n are attributes R = (A 1, A 2, …, A n ) is a relation schema E.g. Customer-schema = (customer-name, customer-street, customer-city) r(R) is a relation on the relation schema R E.g.customer (Customer-schema)

3.6Database System Concepts 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 Jones Smith Curry Lindsay customer-name Main North Park customer-street Harrison Rye Pittsfield customer-city customer attributes (or columns) tuples (or rows)

3.7Database System Concepts Relations are Unordered Order of tuples is irrelevant (tuples may be stored in an arbitrary order) E.g. account relation with unordered tuples

3.8Database System Concepts 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 E.g.: 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. two customers own an account)  the need for null values (e.g. represent a customer without an account)

3.9Database System Concepts The customer Relation

3.10Database System Concepts The depositor Relation

3.11Database System Concepts E-R Diagram for the Banking Enterprise

3.12Database System Concepts 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)  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. K is a candidate key if K is minimal Example: {customer-name} is a candidate key for Customer, since it is a superkey (assuming no two customers can possibly have the same name), and no subset of it is a superkey.

3.13Database System Concepts Determining Keys from E-R Sets Strong entity set. The primary key of the entity set becomes the primary key of the relation. Weak entity set. The primary key of the relation consists of the union of the primary key of the strong entity set and the discriminator of the weak entity set. Relationship set. The union of the primary keys of the related entity sets becomes a super key of the relation.  For binary many-to-one relationship sets, the primary key of the “many” entity set becomes the relation’s primary key.  For one-to-one relationship sets, the relation’s primary key can be that of either entity set.  For many-to-many relationship sets, the union of the primary keys becomes the relation’s primary key

3.14Database System Concepts Schema Diagram for the Banking Enterprise

3.15Database System Concepts Query Languages Language in which user requests information from the database. Categories of languages  procedural  non-procedural “Pure” languages:  Relational Algebra  Tuple Relational Calculus  Domain Relational Calculus Pure languages form underlying basis of query languages that people use.

3.16Database 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.

3.17Database System Concepts Select Operation – Example Relation r ABCD    A=B ^ D > 5 (r) ABCD  

3.18Database 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)

3.19Database System Concepts Project Operation – Example Relation r: ABC  AC  = AC   A,C (r)

3.20Database 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)

3.21Database System Concepts Union Operation – Example Relations r, s: r  s: AB  AB  2323 r s AB 

3.22Database 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)

3.23Database System Concepts Set Difference Operation – Example Relations r, s: r – s : AB  AB  2323 r s AB  1111

3.24Database 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

3.25Database System Concepts Cartesian-Product Operation-Example Relations r, s: r x s: AB  1212 AB  CD  E aabbaabbaabbaabb CD  E aabbaabb r s

3.26Database 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.

3.27Database 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

3.28Database 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.

3.29Database 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)

3.30Database System Concepts Example Queries Find all loans of over $1200 Find the loan number for each loan of an amount greater than $1200  amount > 1200 (loan)  loan-number (  amount > 1200 (loan))

3.31Database System Concepts Example Queries Find the names of all customers who have a loan, an account, or both, from the bank Find the names of all customers who have a loan and an account at bank.  customer-name (borrower)   customer-name (depositor)  customer-name (borrower)   customer-name (depositor)

3.32Database System Concepts Example Queries Find the names of all customers who have a loan at the Perryridge branch. 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 )  customer-name (  branch-name=“Perryridge ” (  borrower.loan-number = loan.loan-number (borrower x loan)))

3.33Database System Concepts Example Queries Find the names of all customers who have a loan at the Perryridge branch.  Query 2  customer-name (  loan.loan-number = borrower.loan-number ( (  branch-name = “Perryridge” (loan)) x borrower))  Query 1  customer-name (  branch-name = “Perryridge” (  borrower.loan-number = loan.loan-number (borrower x loan)))

3.34Database 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)))

3.35Database System Concepts 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 E 1 and E 2 be relational-algebra expressions; the following are all relational-algebra expressions:  E 1  E 2  E 1 - E 2  E 1 x E 2   p (E 1 ), P is a predicate on attributes in E 1   s (E 1 ), S is a list consisting of some of the attributes in E 1   x (E 1 ), x is the new name for the result of E 1

3.36Database 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

3.37Database 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)

3.38Database System Concepts Set-Intersection Operation - Example Relation r, s: r  s A B   2323 r s  2

3.39Database System Concepts Notation: r s Natural-Join Operation 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 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, add a tuple t 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))

3.40Database System Concepts Natural Join Operation – Example Relations r, s: AB  CD  aababaabab B D aaabbaaabb E  r AB  CD  aaaabaaaab E  s r s

3.41Database 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

3.42Database System Concepts Division Operation – Example Relations r, s: r  s:r  s: A B  1212 AB  r s

3.43Database 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

3.44Database 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.

3.45Database 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.

3.46Database System Concepts Find all customers who have an account at all branches located in Brooklyn city. Example Queries  customer-name, branch-name (depositor account)   branch-name (  branch-city = “Brooklyn” (branch))

3.47Database System Concepts Extended Relational-Algebra-Operations Generalized Projection Outer Join Aggregate Functions

3.48Database 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)

3.49Database 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 G1, G2, …, Gn g F1( A1 ), F2( A2 ),…, Fn( An ) (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

3.50Database System Concepts Aggregate Operation – Example Relation r: AB   C g sum(c) (r) sum-C 27

3.51Database 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

3.52Database 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)

3.53Database 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

3.54Database System Concepts Outer Join – Example Relation loan Relation borrower customer-nameloan-number Jones Smith Hayes L-170 L-230 L loan-numberamount L-170 L-230 L-260 branch-name Downtown Redwood Perryridge

3.55Database System Concepts Outer Join – Example Inner Join loan Borrower loan-numberamount L-170 L customer-name Jones Smith branch-name Downtown Redwood Jones Smith null loan-numberamount L-170 L-230 L customer-namebranch-name Downtown Redwood Perryridge Left Outer Join loan Borrower

3.56Database System Concepts Outer Join – Example Right Outer Join loan borrower Full Outer Join loan-numberamount L-170 L-230 L null customer-name Jones Smith Hayes branch-name Downtown Redwood null loan-numberamount L-170 L-230 L-260 L null customer-name Jones Smith null Hayes branch-name Downtown Redwood Perryridge null

3.57Database 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

3.58Database System Concepts Result of  branch-name = “Perryridge” (loan)

3.59Database System Concepts Loan Number and the Amount of the Loan

3.60Database System Concepts Names of All Customers Who Have Either a Loan or an Account

3.61Database System Concepts Customers With An Account But No Loan

3.62Database System Concepts Result of borrower  loan

3.63Database System Concepts Result of  branch-name = “Perryridge” (borrower  loan)

3.64Database System Concepts Result of  customer-name

3.65Database System Concepts Result of the Subexpression

3.66Database System Concepts Largest Account Balance in the Bank

3.67Database System Concepts Customers Who Live on the Same Street and In the Same City as Smith

3.68Database System Concepts Customers With Both an Account and a Loan at the Bank

3.69Database System Concepts Result of  customer-name, loan-number, amount (borrower loan)

3.70Database System Concepts Result of  branch-name (  customer-city = “Harrison” ( customer account depositor))

3.71Database System Concepts Result of  branch-name (  branch-city = “Brooklyn” (branch))

3.72Database System Concepts Result of  customer-name, branch-name (depositor account)

3.73Database System Concepts The credit-info Relation

3.74Database System Concepts Result of  customer-name, (limit – credit-balance) as credit-available (credit-info).

3.75Database System Concepts The pt-works Relation

3.76Database System Concepts The pt-works Relation After Grouping

3.77Database System Concepts Result of branch-name  sum(salary) (pt-works)

3.78Database System Concepts Result of branch-name  sum salary, max(salary) as max-salary (pt-works)

3.79Database System Concepts The employee and ft-works Relations

3.80Database System Concepts The Result of employee ft-works

3.81Database System Concepts The Result of employee ft-works

3.82Database System Concepts Result of employee ft-works

3.83Database System Concepts Result of employee ft-works

3.84Database System Concepts Tuples Inserted Into loan and borrower

3.85Database System Concepts Names of All Customers Who Have a Loan at the Perryridge Branch

3.86Database System Concepts E-R Diagram

3.87Database System Concepts The branch Relation

3.88Database System Concepts The loan Relation

3.89Database System Concepts The borrower Relation