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**Chapter 16: Relational Database Design and Further Dependencies**

TA: Zhe Jiang Ref: Elmasri, Navathe, Fundamentals of Database Systems, 6th, Addison Wesley, ISBN-10:

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**Outline Big picture & motivation**

Simple case algorithm (part of ) Formal algorithm Basic concepts (16.1): General case algorithm (16.3.3)

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**Database Design Phases**

Big Picture: Database Design Phases ER-Diagram Relational Tables Which choice is good? How to guarantee it? Formal Norm Theory

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**Motivation We have: Universal relational schema U(A1,A2, … An).**

A set of functional dependencies (FDs) from domain knowledge. Question: How do we decompose U into sub-relations, so as to satisfy 3NF?

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**Simple Case Decomposition Algorithm**

Motivation: Decompose universal relational schema into sub relations which satisfy 3NF Properties: Preserve dependencies (nonlossy design) Non-additive join property (no spurious tuples) Resulting relational schemas are in 3NF Problem Definition: Input: Universal Relation R and a set of functional dependencies F on the attributes of R Output: Sub-relations, FDs. Constraint: the three properties above

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**Simple Case Decomposition Algorithm**

Suppose the FD set given is already “good” minimal cover (defined later) Approach: For each LHS X in F, create a relation schema in D {X U {A1} U {A2} … U {Ak} }.where XAi only dependency with X as LHS. If none of the relation schemas in D contains a key of R, create one relation with key. (How? Introduce later) Eliminate redundant relations.

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**Simple Case Decomposition Algorithm**

Exercise: Universal relation FD: {PLC, LCAP, AC} Q: Does it satisfy 1NF, 2NF, 3NF? Q: How to decompose the relation to satisfy 3NF? Solution: R1(P,L,C); R2(L,C,A,P); R3(A,C) Already contains key. Remove redundant relations R1 and R3, final answer is R2(L,C,A,P).

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**General Case Decomposition Algorithm**

New info: Transform the given FD set into minimal cover New info: If no key exists, find key of U, then create a relation contain key We will introduce some basic concepts, then formal algorithm

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**Basic Concept Inference rules: One FD could infer another**

trivial: IR1: IR1 (reflexive rule) If X Y, then X Y. non-trivial: IR2-IR4 {XY} |= XZYZ {XY, YZ} |= XZ {XYZ} |=XY Closure of set of dependencies Closure of F: F+, set of all FDs could be inferred. Use IR1 to IR3;

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**Basic Concepts Closure of left-hand-side under dependency set**

Algorithm 16.1 Start: X+={X} Grow X+ with new attributes determined by elements in X+ Repeat 2 until can’t grow any more. Exercise: Given: F={XYZ, XW, WU, YV}, U(X,Y,Z,W,U,V) Find: X+ ?

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**Basic Concepts Equivalence of functional dependencies sets Definition**

Cover: F covers E if F+ contains E. Equivalent FD sets: Algorithm Check if all left-hand-sides’ closures are same Minimal Cover of dependency set F definition: Can’t find subset that is equivalent to F

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**Basic Concept Minimal Cover of dependency set F Example:**

break down right-hand-side, X{A1,A2,…An} to XA1, XA2, …XAn Try reduce size of LHS X in F, e.g. changing X into {X-B} still equivalent to F? Try reduce unnecessary FD in F, e.g. remove XA in F, if result still equivalent to F. Example: F={PLCA, LCAP, AC} What is “minimal cover” of F?

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**Basic Concepts Algorithm to find key of R&F Example: Start with K=R.**

Find A in R such that (K-A)+ contain all attributes. Repeat until size of K is as small as possible Example: U(Emp_ssn, Pno, Esal, Ephone, Dno, Pname, Plocation) F={Emp_ssnEsal, Ephone, Dno; PnoPname, plocation}; What is the key?

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**Decomposition Algorithm: Exercise**

FD3 FD1: Property_id Lot#, County, Area FD2: Lot#, County Area, Property_id FD3: AreaCounty Simpler Version: F={PLCA, LCAP, AC} What is the minimal cover G? Decompose G

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**Decomposition Algorithm Example**

Simpler Version: F={PLCA, LCAP, AC} First Case: Minimal cover GX: F: {PL, PC, PA, LCA, LCP,AC} Minimal cover GX: {PLC, LCAP, AC} Design X: 3. R1(P,L,C), R2(L,C,A,P), and R3(A,C) 4. R2(L,C,A,P)

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**Decomposition Algorithm Example**

Simpler Version: F={PLCA, LCAP, AC} Second Case: Minimal cover GX: F: {PL, PC, PA, LCA, LCP,AC} Minimal cover GX: {PLA, LCP, AC} Design Y: 3. S1(P,A,L), S2(L,C,P), and S3(A,C) 4. No redundant relations.

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**Exercise Given: Universal relation U(A,B,C,D,E,F,G,H,I,J)**

Functional dependencies F={ {A,B}{C}, {B,D}{E,F}, {A,D}{G,H}, {A}{I}, {H}{J} }. Decompose it into 3NF?

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