CMSC424, Spring 2005 CMSC424: Database Design Lecture 4

CMSC424, Spring 2005 Review: Relational Data Model Key Abstraction: Relation Given sets: R = {1, 2, 3}, S = {3, 4} R  S = { (1, 3), (1, 4), (2, 3), (2, 4), (3, 3), (3, 4) } A relation on R, S is any subset (  ) of R  S (e.g: { (1, 4), (3, 4)}) Mathematical relations Account  Branches  Accounts  Balances { (Downtown, A-101,500), (Brighton, A-201, 900), (Brighton, A-217, 500) } Database relations Given attribute domains Branches = { Downtown, Brighton, … } Accounts = { A-101, A-201, A-217, … } Balances = R bnameacct_nobalance Downtown Brighton A-101 A-201 A

CMSC424, Spring 2005 Review: Terms and Definitions 1.Tables = Relations 2.Columns = Attributes 3.Rows = Tuples 4.Relation Schema (or Schema) A list of attributes and their domains We will require the domains to be atomic E.g. account(account-number, branch-name, balance) 5.Relation Instance A particular instantiation of a relation with actual values Will change with time

CMSC424, Spring 2005 Bank Database: Schema Account bnameacct_nobalance Depositor cnameacct_no Customer cnamecstreetccity Branch bnamebcityassets Borrower cnamelno Loan bnamelnoamt

CMSC424, Spring 2005 Bank Database: An Instance Account bnameacct_nobalance Downtown Mianus Perry R.H. Brighton Redwood Brighton A-101 A-215 A-102 A-305 A-201 A-222 A Depositor cnameacct_no Johnson Smith Hayes Turner Johnson Jones Lindsay A-101 A-215 A-102 A-305 A-201 A-217 A-222 Customer cnamecstreetccity Jones Smith Hayes Curry Lindsay Turner Williams Adams Johnson Glenn Brooks Green Main North Main North Park Putnam Nassau Spring Alma Sand Hill Senator Walnut Harrison Rye Harrison Rye Pittsfield Stanford Princeton Pittsfield Palo Alto Woodside Brooklyn Stanford Branch bnamebcityassets Downtown Redwood Perry Mianus R.H. Pownel N. Town Brighton Brooklyn Palo Alto Horseneck Bennington Rye Brooklyn 9M 2.1M 1.7M 0.4M 8M 0.3M 3.7M 7.1M Borrower cnamelno Jones Smith Hayes Jackson Curry Smith Williams Adams L-17 L-23 L-15 L-14 L-93 L-11 L-17 L-16 Loan bnamelnoamt Downtown Redwood Perry Downtown Mianus R.H. Perry L-17 L-23 L-15 L-14 L-93 L-11 L

CMSC424, Spring 2005 Review: Keys and Relations 1. Superkeys set of attributes of table for which every row has distinct set of values 2. Candidate keys “minimal” superkeys 3. Primary keys DBA-chosen candidate keys As in the E/R Model: Act as Integrity Constraints i.e., guard against illegal/invalid instance of given schema e.g., Branch = (bname, bcity, assets)  bnamebcityassets Brighton Brooklyn Boston 5M 3M Invalid!!

CMSC424, Spring 2005 More on Keys Determining Primary Keys If relation schema derived from E-R diagrams, we can determine the primary keys using the original entity and relationship sets Otherwise, same way we do it for E-R diagrams Find candidate keys (minimal sets of attributes that can uniquely identify a tuple) Designate one of them to be primary key Foreign Keys If a relation schema includes the primary key of another relation schema, that attribute is called the foreign key

CMSC424, Spring 2005 Schema Diagram for the Banking Enterprise

CMSC424, Spring 2005 Relational Query Languages Theoretical QL’s give semantics to Practical QL’s Recall: Query = “Retrieval Program” Theoretical: 1.Relational Algebra 2.Relational Calculus a.Tuple Relational Calculus (TRC) b.Domain Relational Calculus (DRC) Practical: 1.SQL (originally: SEQUEL from System R) 2.Quel (used in Ingres) 3.Datalog (Prolog-like – used in research lab systems) Language Examples:

CMSC424, Spring 2005 Relational Algebra Basic Operators 1.select ( σ ) 2.project (  ) 3.union (  ) 4.set difference ( – ) 5.cartesian product (  ) 6.rename ( ρ ) Relational Operator Relation

CMSC424, Spring 2005 Select ( σ ) Notation: σ predicate (Relation) Relation: Can be name of table, or another query Predicate: 1. Simple attribute 1 = attribute 2 attribute = constant value (also: ≠,, ≤, ≥) 2. Complex predicate AND predicate predicate OR predicate NOT predicate

CMSC424, Spring 2005 Select ( σ ) Examples: bnamebcityassets Downtown Brighton Brooklyn 9M 7.1M bnamebcityassets DowntownBrooklyn9M σ bcity = “Brooklyn” (branch) = σ assets > 8M (σ bcity = “Brooklyn” ( branch )) = Notation: σ predicate (Relation)

CMSC424, Spring 2005 Project (  ) Notation:  A1, …, An (Relation) Each A i an attribute Idea:  selects columns (vs. σ which selects rows)  cstreet, ccity (customer) = cstreetccity Main North Park Putnam Nassau Spring Alma Sand Hill Senator Walnut Harrison Rye Pittsfield Stanford Princeton Pittsfield Palo Alto Woodside Brooklyn Stanford Examples:

CMSC424, Spring 2005 Project (  ) Examples: Notation:  A1, …, An (Relation) Each A i an attribute Idea:  selects columns (vs. σ which selects rows)   bcity ( σ assets > 5M (branch) ) = bcity Brooklyn Horseneck

CMSC424, Spring 2005 Union (  ) Notation: Relation 1  Relation 2 Example: (   cname (depositor))  (  cname (borrower)) = cname Johnson Smith Hayes Turner Jones Lindsay Jackson Curry Williams Adams R  S valid only if: 1.R, S have same number of columns (arity) 2.R, S corresponding columns have same domain (compatibility)

CMSC424, Spring 2005 Set Difference ( – ) bnamelnoamount Downtown Redwood Perry Downtown Perry L-17 L-23 L-15 L-14 L Notation: Relation 1 - Relation 2 R - S valid only if: 1.R, S have same number of columns (arity) 2.R, S corresponding columns have same domain (compatibility) Example: (  bname ( σ amount ≥ 1000 (loan))) – (  bname ( σ balance < 800 (account))) = bnameacct_nobalance Mianus Brighton Redwood Brighton A-215 A-201 A-222 A

CMSC424, Spring 2005 Set Difference ( – ) bnamelnoamount Downtown Redwood Perry Downtown Perry L-17 L-23 L-15 L-14 L Notation: Relation 1 - Relation 2 R - S valid only if: 1.R, S have same number of columns (arity) 2.R, S corresponding columns have same domain (compatibility) Example: (  bname ( σ amount ≥ 1000 (loan))) – (  bname ( σ balance < 800 (account))) = bnameacct_nobalance Mianus Brighton Redwood Brighton A-215 A-201 A-222 A – = bname Downtown Perry

CMSC424, Spring 2005 Cartesian Product (  ) Notation: Relation 1  Relation 2 Example: R  S like cross product for mathematical relations: every tuple of R appended to every tuple of S depositor  borrower = depositor. cname acct_noborrower. cname lno Johnson Smith … A-101 A-215 … Jones Smith Hayes Jackson Curry Smith Williams Adams Jones … L-17 L-23 L-15 L-14 L-93 L-11 L-17 L-16 L-17 … How many tuples in the result? A: 56

CMSC424, Spring 2005 Rename ( ρ ) Notation:  identifier (Relation) renames a relation, or Notation:  identifier0 (identifier1, …, identifiern) (Relation) renames relation and columns of n-column relation Use: massage relations to make , – valid, or  more readable

CMSC424, Spring 2005 Rename ( ρ ) Notation:  identifier0 (identifier1, …, identifiern) (Relation) renames relation and columns of n-column relation Example:  res (dcname, acctno, bcname, lno) (depositor  borrower) = depositor. cname acct_noborrower. cname lno Johnson Smith … A-101 A-215 … Jones Smith Hayes Jackson Curry Smith Williams Adams Jones … L-17 L-23 L-15 L-14 L-93 L-11 L-17 L-16 L-17 …

CMSC424, Spring 2005 Rename ( ρ ) Notation:  identifier0 (identifier1, …, identifiern) (Relation) renames relation and columns of n-column relation Example:  res (dcname, acctno, bcname, lno) (depositor  borrower) = dcnameacctnobcnamelno Johnson Smith … A-101 A-215 … Jones Smith Hayes Jackson Curry Smith Williams Adams Jones … L-17 L-23 L-15 L-14 L-93 L-11 L-17 L-16 L-17 … res =

CMSC424, Spring 2005 Example Query in RA Determine lno’s for loans that are for an amount that is larger than the amt of some other loan. (i.e. lno’s for all non-minimal loans) Temp 1  … Temp 2  … Temp 1 … … Can do in steps:

CMSC424, Spring 2005 Bank Database: An Instance Account bnameacct_nobalance Downtown Mianus Perry R.H. Brighton Redwood Brighton A-101 A-215 A-102 A-305 A-201 A-222 A Depositor cnameacct_no Johnson Smith Hayes Turner Johnson Jones Lindsay A-101 A-215 A-102 A-305 A-201 A-217 A-222 Customer cnamecstreetccity Jones Smith Hayes Curry Lindsay Turner Williams Adams Johnson Glenn Brooks Green Main North Main North Park Putnam Nassau Spring Alma Sand Hill Senator Walnut Harrison Rye Harrison Rye Pittsfield Stanford Princeton Pittsfield Palo Alto Woodside Brooklyn Stanford Branch bnamebcityassets Downtown Redwood Perry Mianus R.H. Pownel N. Town Brighton Brooklyn Palo Alto Horseneck Bennington Rye Brooklyn 9M 2.1M 1.7M 0.4M 8M 0.3M 3.7M 7.1M Borrower cnamelno Jones Smith Hayes Jackson Curry Smith Williams Adams L-17 L-23 L-15 L-14 L-93 L-11 L-17 L-16 Loan bnamelnoamt Downtown Redwood Perry Downtown Mianus R.H. Perry L-17 L-23 L-15 L-14 L-93 L-11 L

CMSC424, Spring 2005 lnoamt L-17 L-23 L-15 L-14 L-93 L-11 L Example Query in RA 1. Find the base data we need Temp 1   lno,amt (loan) 2. Make a copy of (1) Temp 2  ρ Temp2 (lno2,amt2) (Temp 1 ) lno2amt2 L-17 L-23 L-15 L-14 L-93 L-11 L

CMSC424, Spring 2005 lnoamtlno2amt2 L-17 … L-17 L-23 … L-23 … 1000 … … 2000 … L-17 L-23 … L-16 L-17 L-23 … L-16 … … … 1300 … 3. Take the cartesian product of 1 and 2 Temp 3  Temp 1  Temp 2 Example Query in RA

CMSC424, Spring 2005  lno ( σ amt > amt2 (  lno,amt (loan)  (ρ Temp2 (lno2,amt2) (  lno,amt (loan))))) 4. Select non-minimal loans Temp 4  σ amt > amt2 (Temp 3 ) 5. Project on lno Result   lno (Temp 4 ) Example Query in RA … or, if you prefer…

CMSC424, Spring 2005 What we learned so far… Relational Algebra Operators 1.Select 2.Project 3.Set Union 4.Set Difference 5.Cartesian Product 6.Rename These are called fundamental operations

CMSC424, Spring 2005 Formal Definition Basic expression A relation in the database A constant relation e.g. {(A-101, Downtown, 500), (A-215, Mianus, 700)…} Let E 1 and E 2 be two relational-algebra expressions, then the following are also: 1.σ P (E 1 ), where P is a predicate on attributes in E 1 2.p S (E 1 )  where S is a list containing some attributes in E 1 3.E 1  E 2, 4.E 1 – E 2 5.E 1  E 2 6.ρ x (E 1 ), where x is the new name for the result of E 1

CMSC424, Spring 2005 Relational Algebra Redundant Operators 4. Update (  ) (we’ve already been using) 2. Division ( ) 1.Natural Join ( ) 3. Outer Joins ( ) Redundant: Above can be expressed in terms of minimal RA  e.g. depositor borrower = π …(σ…(depositor  ρ…(borrower))) Added as convenience

CMSC424, Spring 2005 Natural Join Idea: combines ρ, , σ ABCD αααβαααβ EBD ‘a’ ‘b’ ‘c’ ααββααββ r s ABCDE αααββαααββ ‘a’ ‘b’ ‘c’ = Relation 1 Relation 2 Notation: π cname,acct_no,lno (σ cname=cname2 (depositor  ρ t(cname2,lno) (borrower))) ≡ depositor borrower

CMSC424, Spring 2005 Division AB αααβγγγγδδαααβγγγγδδ B 1212 r s A αδαδ = Query: Find values for A in r which have corresponding B values for all B values in s Relation 1 Relation 2 Notation: Idea: expresses “for all” queries

CMSC424, Spring 2005 Division AB αααβγγγγδδαααβγγγγδδ B 1212 r s A αδαδ = 17  3 = 5 The largest value of i such that: i  3 ≤ 17 t Relational Division The largest value of t such that: ( t  s  r ) Another way to look at it: and 

CMSC424, Spring 2005 ABCDE αααββγγγαααββγγγ aaaaaaaaaaaaaaaa αγγγγγγβαγγγγγγβ aabababbaabababb DE abab 1111 r s ABC αγαγ aaaa γγγγ = t ? Division A More Complex Example