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Midterm 2 Revision Prof. Sin-Min Lee Department of Computer Science San Jose State University

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©Silberschatz, Korth and Sudarshan3.2Database System Concepts 1. Know the difference between a database and a DBMS Functions/advantages and disadvantages of a DBMS 2.Understand the meaning of all of the E-R symbols. 3.Know the basis of the mathematical relation and the properties of a relation. Understand and recognize symbols for Selection, projection, Cartesian product, union and set difference. Understand the difference between an inner join and an outerjoin 4.Know the characteristics of superkey, candidate key, primary key, and foreign key. 5.Know the rules of relational integrity and referential integrity. 6. Be able to recognize and read relational algebra statements with the primary operators. 7.Be able to recognized simple relational calculus statements (like the ones used in class) and understand the difference between the algebra and calculus. Materials cover in Exam.

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©Silberschatz, Korth and Sudarshan3.3Database System Concepts Multiple Choice Problems

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©Silberschatz, Korth and Sudarshan3.4Database System Concepts

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©Silberschatz, Korth and Sudarshan3.5Database System Concepts

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©Silberschatz, Korth and Sudarshan3.6Database System Concepts E-R Diagram for the Banking Enterprise

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©Silberschatz, Korth and Sudarshan3.7Database 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

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©Silberschatz, Korth and Sudarshan3.8Database System Concepts Chapter 3: Relational Model Structure of Relational Databases Relational Algebra Tuple Relational Calculus Domain Relational Calculus Extended Relational-Algebra-Operations

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©Silberschatz, Korth and Sudarshan3.9Database 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 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

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©Silberschatz, Korth and Sudarshan3.10Database 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

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©Silberschatz, Korth and Sudarshan3.11Database 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)

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©Silberschatz, Korth and Sudarshan3.12Database 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 tuples

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©Silberschatz, Korth and Sudarshan3.13Database 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

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©Silberschatz, Korth and Sudarshan3.14Database 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.

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©Silberschatz, Korth and Sudarshan3.15Database 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.

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©Silberschatz, Korth and Sudarshan3.16Database 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.

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©Silberschatz, Korth and Sudarshan3.17Database System Concepts Select Operation – Example Relation r ABCD 1 5 12 23 7 3 10 A=B ^ D > 5 (r) ABCD 1 23 7 10

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©Silberschatz, Korth and Sudarshan3.18Database 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)

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©Silberschatz, Korth and Sudarshan3.19Database System Concepts Project Operation – Example Relation r: ABC 10 20 30 40 11121112 AC 11121112 = AC 112112 A,C (r)

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©Silberschatz, Korth and Sudarshan3.20Database 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)

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©Silberschatz, Korth and Sudarshan3.21Database System Concepts Union Operation – Example Relations r, s: r s: AB 121121 AB 2323 r s AB 12131213

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©Silberschatz, Korth and Sudarshan3.22Database 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)

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©Silberschatz, Korth and Sudarshan3.23Database System Concepts Set Difference Operation – Example Relations r, s: r – s : AB 121121 AB 2323 r s AB 1111

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©Silberschatz, Korth and Sudarshan3.24Database 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

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©Silberschatz, Korth and Sudarshan3.25Database System Concepts Cartesian-Product Operation-Example Relations r, s: r x s: AB 1212 AB 1111222211112222 CD 10 19 20 10 20 10 E aabbaabbaabbaabb CD 20 10 E aabbaabb r s

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©Silberschatz, Korth and Sudarshan3.26Database 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.

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©Silberschatz, Korth and Sudarshan3.27Database 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 1111222211112222 CD 10 19 20 10 20 10 E aabbaabbaabbaabb ABCDE 122122 20 aabaab

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©Silberschatz, Korth and Sudarshan3.28Database 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.

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©Silberschatz, Korth and Sudarshan3.29Database 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)

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©Silberschatz, Korth and Sudarshan3.30Database 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))

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©Silberschatz, Korth and Sudarshan3.31Database 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)

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©Silberschatz, Korth and Sudarshan3.32Database 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 )

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©Silberschatz, Korth and Sudarshan3.33Database 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) )

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©Silberschatz, Korth and Sudarshan3.34Database 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)))

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©Silberschatz, Korth and Sudarshan3.35Database 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

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©Silberschatz, Korth and Sudarshan3.36Database 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)

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©Silberschatz, Korth and Sudarshan3.37Database System Concepts Set-Intersection Operation - Example Relation r, s: r s A B 121121 2323 r s 2

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©Silberschatz, Korth and Sudarshan3.38Database 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))

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©Silberschatz, Korth and Sudarshan3.39Database System Concepts Natural Join Operation – Example Relations r, s: AB 1241212412 CD aababaabab B 1312313123 D aaabbaaabb E r AB 1111211112 CD aaaabaaaab E s r s

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©Silberschatz, Korth and Sudarshan3.40Database 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

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©Silberschatz, Korth and Sudarshan3.41Database System Concepts Division Operation – Example Relations r, s: r s:r s: A B 1212 AB 1231113461212311134612 r s

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©Silberschatz, Korth and Sudarshan3.42Database System Concepts Another Division Example AB aaaaaaaaaaaaaaaa CD aabababbaabababb E 1111311111113111 Relations r, s: r s:r s: D abab E 1111 AB aaaa C r s

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©Silberschatz, Korth and Sudarshan3.43Database 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.

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©Silberschatz, Korth and Sudarshan3.44Database 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.

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©Silberschatz, Korth and Sudarshan3.45Database 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”)})

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©Silberschatz, Korth and Sudarshan3.46Database 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

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©Silberschatz, Korth and Sudarshan3.47Database System Concepts Extended Relational-Algebra-Operations Generalized Projection Outer Join Aggregate Functions

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©Silberschatz, Korth and Sudarshan3.48Database 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)

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©Silberschatz, Korth and Sudarshan3.49Database 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 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

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©Silberschatz, Korth and Sudarshan3.50Database System Concepts Aggregate Operation – Example Relation r: AB C 7 3 10 g sum(c) (r) sum-C 27

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©Silberschatz, Korth and Sudarshan3.51Database 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-222 400 900 750 700 branch-namebalance Perryridge Brighton Redwood 1300 1500 700

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©Silberschatz, Korth and Sudarshan3.52Database 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)

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©Silberschatz, Korth and Sudarshan3.53Database 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

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©Silberschatz, Korth and Sudarshan3.54Database System Concepts Outer Join – Example Relation loan loan-numberamount L-170 L-230 L-260 3000 4000 1700 Relation borrower customer-nameloan-number Jones Smith Hayes L-170 L-230 L-155 branch-name Downtown Redwood Perryridge

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©Silberschatz, Korth and Sudarshan3.55Database System Concepts Outer Join – Example Inner Join loan Borrower loan borrower Left Outer Join loan-numberamount L-170 L-230 3000 4000 customer-name Jones Smith branch-name Downtown Redwood loan-numberamount L-170 L-230 L-260 3000 4000 1700 customer-name Jones Smith null branch-name Downtown Redwood Perryridge

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©Silberschatz, Korth and Sudarshan3.56Database System Concepts Outer Join – Example Right Outer Join loan borrower loan-numberamount L-170 L-230 L-155 3000 4000 null customer-name Jones Smith Hayes loan-numberamount L-170 L-230 L-260 L-155 3000 4000 1700 null customer-name Jones Smith null Hayes loan borrower Full Outer Join branch-name Downtown Redwood null branch-name Downtown Redwood Perryridge null

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©Silberschatz, Korth and Sudarshan3.57Database 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

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©Silberschatz, Korth and Sudarshan3.58Database System Concepts

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©Silberschatz, Korth and Sudarshan3.59Database 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

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©Silberschatz, Korth and Sudarshan3.60Database 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.

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©Silberschatz, Korth and Sudarshan3.61Database 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.

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©Silberschatz, Korth and Sudarshan3.62Database 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

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©Silberschatz, Korth and Sudarshan3.63Database 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.

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©Silberschatz, Korth and Sudarshan3.64Database 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 )

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©Silberschatz, Korth and Sudarshan3.65Database 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

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©Silberschatz, Korth and Sudarshan3.66Database 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 10000 (account)) AN, BN, BAL * 1.05 ( BAL 10000 (account))

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©Silberschatz, Korth and Sudarshan3.67Database System Concepts

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©Silberschatz, Korth and Sudarshan3.68Database System Concepts R(A B C D) 1 2 2 1 Functional Dependency Graph 2 1 2 3 A B C D 1 2 4 2 3 1 2 1 3 1 1 4

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