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CSCI 4333 Database Design and Implementation Review for Midterm Exam II Xiang Lian The University of Texas – Pan American Edinburg, TX

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Review Chapters 5 ~ 7 in your textbook Lecture slides In-class exercises Assignments Projects 2

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Review (cont'd) Question Types – Q/A Relational algebra SQL Normalization theory – 3 axioms, FD closure, attribute closure, BCNF, 3NF, minimal cover, lossless decomposition, dependency preserving 5 Questions (100 points) + 1 Bonus Question (20 extra points) 3

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Chapter 5 Relational Algebra and SQL Relational algebra – Select, project, set operators, union, cartesian product, (natural) join, division SQL – SQL for operators above – Aggregates – Group by … Having – Order by 4

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5 Select Operator Produce table containing subset of rows of argument table satisfying condition condition (relation) Example: Person Person Person Hobby=‘stamps’ ( Person ) 1123 John 123 Main stamps 1123 John 123 Main coins 5556 Mary 7 Lake Dr hiking 9876 Bart 5 Pine St stamps 1123 John 123 Main stamps 9876 Bart 5 Pine St stamps Id Name Address Hobby

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6 Project Operator Produces table containing subset of columns of argument table attribute list (relation) Example: PersonPerson Person Name,Hobby (Person) 1123 John 123 Main stamps 1123 John 123 Main coins 5556 Mary 7 Lake Dr hiking 9876 Bart 5 Pine St stamps John stamps John coins Mary hiking Bart stamps Id Name Address Hobby Name Hobby

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7 Set Operators Relation is a set of tuples, so set operations should apply: , , (set difference) Result of combining two relations with a set operator is a relation => all its elements must be tuples having same structure union compatible relations Hence, scope of set operations limited to union compatible relations

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8 Union Compatible Relations union compatible Two relations are union compatible if – Both have same number of columns – Names of attributes are the same in both – Attributes with the same name in both relations have the same domain unionintersectionsetdifference Union compatible relations can be combined using union, intersection, and set difference

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9 Cartesian Product RSRS R S If R and S are two relations, R S is the set of all concatenated tuples, where x is a tuple in R and y is a tuple in S – RS – R and S need not be union compatible RS R S is expensive to compute: – Factor of two in the size of each row – Quadratic in the number of rows A B C D A B C D x1 x2 y1 y2 x1 x2 y1 y2 x3 x4 y3 y4 x1 x2 y3 y4 x3 x4 y1 y2 RS R S x3 x4 y3 y4 RS R S

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10 Derived Operation: Join generalthetajoin A (general or theta) join of R and S is the expression R join-condition S where join-condition is a conjunction of terms: A i oper B i in which A i is an attribute of R; B i is an attribute of S; and oper is one of =,, , . The meaning is: join-condition ´ (R S) where join-condition and join-condition ´ are the same, except for possible renamings of attributes (next)

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11 Natural Join Special case of equijoin: – join condition equates all and only those attributes with the same name (condition doesn’t have to be explicitly stated) – duplicate columns eliminated from the result Transcript Transcript (StudId, CrsCode, Sem, Grade) Teaching ( Teaching (ProfId, CrsCode, Sem) Transcript Teaching Teaching = StudId, Transcript.CrsCode, Transcript.Sem, Grade, ProfId Transcript ( Transcript Sem Teaching CrsCode=CrsCode AND Sem=Sem Teaching ) [ StudId, CrsCode, Sem, Grade, ProfId ]

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12 Division Goal: Produce the tuples in one relation, r, that match all tuples in another relation, s – r – r (A 1, …A n, B 1, …B m ) – s – s (B 1 …B m ) – rs s r – r/s, with attributes A 1, …A n, is the set of all tuples such that for every tuple in s, is in r Can be expressed in terms of projection, set difference, and cross-product

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13 Set Operators SQL provides UNION, EXCEPT (set difference), and INTERSECT for union compatible tables Example: Find all professors in the CS Department and all professors that have taught CS courses ( SELECT P.Name ProfessorTeaching FROM Professor P, Teaching T WHERE P.Id=T.ProfId AND T.CrsCode LIKE ‘CS%’) UNION (SELECT P.Name Professor FROM Professor P WHERE P.DeptId = ‘CS’)

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14 Division in SQL Query type: Find the subset of items in one set that are related to all items in another set Example: Find professors who taught courses in all departments – Why does this involve division? ProfId DeptIdDeptId All department Ids Contains row if professor p taught a course in department d ProfId,DeptId (Teaching Course) / DeptId (Department)

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15 Aggregates Functions that operate on sets: – COUNT, SUM, AVG, MAX, MIN Produce numbers (not tables) Not part of relational algebra (but not hard to add) SELECT COUNT(*) Professor FROM Professor P SELECT MAX (Salary) Employee FROM Employee E

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16 Grouping But how do we compute the number of courses taught in S2000 per professor? – Strategy 1: Fire off a separate query for each professor: SELECT COUNT( T.CrsCode ) Teaching FROM Teaching T WHERE T.Semester = ‘S2000’ AND T.ProfId = Cumbersome What if the number of professors changes? Add another query? grouping operator – Strategy 2: define a special grouping operator: SELECT T.ProfId, COUNT( T.CrsCode ) Teaching FROM Teaching T WHERE T.Semester = ‘S2000’ GROUP BY T.ProfId

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17 HAVING Clause Eliminates unwanted groups (analogous to WHERE clause, but works on groups instead of individual tuples ) HAVING condition is constructed from attributes of GROUP BY list and aggregates on attributes not in that list SELECT T.StudId, AVG (T.Grade) AS CumGpa, COUNT (*) AS NumCrs Transcript FROM Transcript T WHERE T.CrsCode LIKE ‘CS%’ GROUP BY T.StudId HAVING AVG (T.Grade) > 3.5

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18 ORDER BY Clause Causes rows to be output in a specified order SELECT T.StudId, COUNT (*) AS NumCrs, AVG (T.Grade) AS CumGpa Transcript FROM Transcript T WHERE T.CrsCode LIKE ‘CS%’ GROUP BY T.StudId HAVING AVG (T.Grade) > 3.5 ORDER BY DESC CumGpa, ASC StudId Descending Ascending

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Chapter 6 Relational Normalization Theory Redundancy in the schema – Update, deletion, insertion anomalies – Solution: decomposition Normalization theory – Functional dependencies – FD closure – Attribute closure 19

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Chapter 6 Relational Normalization Theory (cont'd) – BCNF What are two conditions of BCNF? BCNF decomposition algorithm – 3NF What are 3 conditions of 3NF? How to calculate the minimal cover? 3NF decomposition algorithm – Lossless decomposition Conditions? R = R 1 R 2 … R n – Dependency preserving Conditions? F + = (F1 F2 F n ) + 20

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21 Redundancy Dependencies between attributes cause redundancy – Ex. All addresses in the same town have the same zip code SSN Name Town Zip 1234 Joe Stony Brook Mary Stony Brook Tom Stony Brook …………………. Redundancy

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22 Anomalies Redundancy leads to anomalies: – Update anomaly: A change in Address must be made in several places – Deletion anomaly: Suppose a person gives up all hobbies. Do we: Set Hobby attribute to null? No, since Hobby is part of key Delete the entire row? No, since we lose other information in the row – Insertion anomaly: Hobby value must be supplied for any inserted row since Hobby is part of key

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23 Decomposition Person Solution: use two relations to store Person information – Person1 – Person1 (SSN, Name, Address) – Hobbies – Hobbies (SSN, Hobby) The decomposition is more general: people without hobbies can now be described No update anomalies: – Name and address stored once – A hobby can be separately supplied or deleted

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24 Functional Dependencies functional dependency Definition: A functional dependency (FD) on a relation schema R is a constraint of the form X Y, where X and Y are subsets of attributes of R. satisfied Definition: An FD X Y is satisfied in an instance r of R if for every pair of tuples, t and s: if t and s agree on all attributes in X then they must agree on all attributes in Y – Key constraint is a special kind of functional dependency: all attributes of relation occur on the right-hand side of the FD: SSN SSN, Name, Address

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25 Armstrong’s Axioms for FDs This is the syntactic way of computing/testing the various properties of FDs Reflexivity: If Y X then X Y (trivial FD) – Name, Address Name Augmentation: If X Y then X Z YZ – If Town Zip then Town, Name Zip, Name Transitivity: If X Y and Y Z then X Z

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26 Generating F + F AB C AB BCD A D AB BD AB BCDE AB CDE D E BCD BCDE Thus, AB BD, AB BCD, AB BCDE, and AB CDE are all elements of F + union aug trans aug decomp

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27 Computation of Attribute Closure X + F closure := X; // since X X + F repeat old := closure; if there is an FD Z V in F such that Z closure and V closure then closure := closure V until old = closure – If T closure then X T is entailed by F

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28 Example: Computation of Attribute Closure AB C (a) A D (b) D E (c) AC B (d) Problem: Compute the attribute closure of AB with respect to the set of FDs : Initially closure = {AB} Using (a) closure = {ABC} Using (b) closure = {ABCD} Using (c) closure = {ABCDE} Solution:

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29 BCNF Definition: A relation schema R is in BCNF if for every FD X Y associated with R either – Y X (i.e., the FD is trivial) or – X is a superkey of R Person1 Example: Person1(SSN, Name, Address) – The only FD is SSN Name, Address Person1 – Since SSN is a key, Person1 is in BCNF

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30 Third Normal Form A relational schema R is in 3NF if for every FD X Y associated with R either: – Y X (i.e., the FD is trivial); or – X is a superkey of R; or – Every A Y is part of some key of R 3NF is weaker than BCNF (every schema that is in BCNF is also in 3NF) BCNF conditions

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31 Lossless Schema Decomposition A decomposition should not lose information lossless A decomposition (R 1,…,R n ) of a schema, R, is lossless if every valid instance, r, of R can be reconstructed from its components: where each r i = Ri (r) r = r 1 r2r2 rnrn ……

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32 Testing for Losslessness A (binary) decomposition of R = (R, F) into R 1 = (R 1, F 1 ) and R 2 = (R 2, F 2 ) is lossless if and only if : – either the FD (R 1 R 2 ) R 1 is in F + – or the FD (R 1 R 2 ) R 2 is in F +

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33 Dependency Preservation Consider a decomposition of R = (R, F) into R 1 = (R 1, F 1 ) and R 2 = (R 2, F 2 ) – An FD X Y of F + is in F i iff X Y R i – An FD, f F + may be in neither F 1, nor F 2, nor even (F 1 F 2 ) + Checking that f is true in r 1 or r 2 is (relatively) easy Checking f in r 1 r 2 is harder – requires a join Ideally: want to check FDs locally, in r 1 and r 2, and have a guarantee that every f F holds in r 1 r 2 dependency preserving The decomposition is dependency preserving iff the sets F and F 1 F 2 are equivalent: F + = (F 1 F 2 ) + – Then checking all FDs in F, as r 1 and r 2 are updated, can be done by checking F 1 in r 1 and F 2 in r 2

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34 BCNF Decomposition Algorithm Input: R = (R; F) Decomp := R while there is S = (S; F ’ ) Decomp and S not in BCNF do Find X Y F ’ that violates BCNF // X isn’t a superkey in S Replace S in Decomp with S 1 = (XY; F 1 ), S 2 = (S - (Y - X); F 2 ) // F 1 = all FDs of F ’ involving only attributes of XY // F 2 = all FDs of F ’ involving only attributes of S - (Y - X) end return Decomp

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35 Third Normal Form Compromise – Not all redundancy removed, but dependency preserving decompositions are always possible (and, of course, lossless) 3NF decomposition is based on a minimal cover

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36 Minimal Cover minimal cover A minimal cover of a set of dependencies, F, is a set of dependencies, U, such that: – U is equivalent to F (F + = U + ) – All FDs in U have the form X A where A is a single attribute – It is not possible to make U smaller (while preserving equivalence) by Deleting an FD Deleting an attribute from an FD (either from LHS or RHS) redundant – FDs and attributes that can be deleted in this way are called redundant

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37 Computing Minimal Cover Example: F = {ABH CK, A D, C E, BGH L, L AD, E L, BH E} step 1: Make RHS of each FD into a single attribute – Algorithm: Use the decomposition inference rule for FDs – Example: L AD replaced by L A, L D ; ABH CK by ABH C, ABH K step 2: Eliminate redundant attributes from LHS. – Algorithm: If FD XB A F (where B is a single attribute) and X A is entailed by F, then B was unnecessary – Example: Can an attribute be deleted from ABH C ? Compute AB + F, AH + F, BH + F. Since C (BH) + F, BH C is entailed by F and A is redundant in ABH C.

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38 Computing Minimal Cover (con’t) step 3: Delete redundant FDs from F – Algorithm: If F – {f} entails f, then f is redundant If f is X A then check if A X + F-{f} – Example: BGH L is entailed by E L, BH E, so it is redundant Note: The order of steps 2 and 3 cannot be interchanged!! See the textbook for a counterexample

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39 Synthesizing a 3NF Schema step 1: Compute a minimal cover, U, of F. The decomposition is based on U, but since U + = F + the same functional dependencies will hold – A minimal cover for F={ABH CK, A D, C E, BGH L, L AD, E L, BH E} is U={BH C, BH K, A D, C E, L A, E L} Starting with a schema R = (R, F)

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40 Synthesizing a 3NF schema (con’t) step 2: Partition U into sets U 1, U 2, … U n such that the LHS of all elements of U i are the same – U 1 = {BH C, BH K}, U 2 = {A D}, U 3 = {C E}, U 4 = {L A}, U 5 = {E L}

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41 Synthesizing a 3NF schema (con’t) step 3: For each U i form schema R i = (R i, U i ), where R i is the set of all attributes mentioned in U i – Each FD of U will be in some R i. Hence the decomposition is dependency preserving – R 1 = (BHCK; BH C, BH K), R 2 = (AD; A D), R 3 = (CE; C E), R 4 = (AL; L A), R 5 = (EL; E L)

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42 Synthesizing a 3NF schema (con’t) step 4: If no R i is a superkey of R, add schema R 0 = (R 0,{}) where R 0 is a key of R. – R 0 = (BGH, {}) R 0 might be needed when not all attributes are necessarily contained in R 1 R 2 … R n – A missing attribute, A, must be part of all keys (since it’s not in any FD of U, deriving a key constraint from U involves the augmentation axiom) R 0 might be needed even if all attributes are accounted for in R 1 R 2 … R n – Example: (ABCD; {A B, C D}). Step 3 decomposition: R 1 = (AB; {A B}), R 2 = (CD; {C D}). Lossy! Need to add (AC; { }), for losslessness – Step 4 guarantees lossless decomposition.

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Chapter 7 Triggers and Active Databases Syntax of trigger – Events – Conditions – Actions 43

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44 Trigger Overview Element of the database schema General form: ON IF THEN – Event- request to execute database operation – Condition - predicate evaluated on database state – Action – execution of procedure that might involve database updates Example: ON updating maximum course enrollment IF number registered > new max enrollment limit THEN deregister students using LIFO policy

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45 Triggers in SQL:1999 Events: INSERT, DELETE, or UPDATE statements or changes to individual rows caused by these statements Condition: Anything that is allowed in a WHERE clause Action: An individual SQL statement or a program written in the language of Procedural Stored Modules (PSM) (which can contain embedded SQL statements)

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46 Triggers in SQL:1999 Consideration: Immediate – Condition can refer to both the state of the affected row or table before and after the event occurs Execution: Immediate – can be before or after the execution of the triggering event – Action of before trigger cannot modify the database Granularity: Both row-level and statement-level

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