Christoph FEickFunctional Dependencies,BCNF,and

Click to edit Master title style,Click to edit Master text styles,Second Level,Third Level,Fourth Level,Fifth Level,Christoph F. Eick:,Functional Dependencies, BCNF, and Normalization,26,Functional Dependencies, BCNF and Normalization,Functional Dependencies (FDs),A,functional dependency,X Y,holds over relation R if, for every allowable instance,r,of R:,t1 r, t2 r,(,t1,) = (,t2,) implies (,t1,) = (,t2,),i.e., given two tuples in,r, if the X values agree, then the Y values must also agree. (X and Y are,sets,of attributes.),An FD is a statement about,all,allowable relations.,Must be identified based on semantics of application.,Given some allowable instance,r1,of R, we can check if it violates some FD,f, but we cannot tell if,f,holds over R!,K is a candidate key for R means that K R,However, K R does not require K to be,minimal,!,Reasoning About FDs,Given some FDs, we can usually infer additional FDs:,ssn did,did lot,implies,ssn lot,An FD,f,is,implied by,a set of FDs,F,if,f,holds whenever all FDs in,F,hold.,=,closure of F,is the set of all FDs that are implied by,F,.,Armstrongs Axioms (X, Y, Z are sets of attributes):,Reflexivity,:,If Y X, then X Y,Augmentation,:,If X Y, then XZ YZ for any Z,Transitivity,:,If X Y and Y Z, then X Z,These are,sound,and,complete,inference rules for FDs!,Reasoning About FDs (Contd.),Couple of additional rules (that follow from AA):,Union,:,If X Y and X Z, then X YZ,Decomposition,:,If X YZ, then X Y and X Z,Example:,Contracts(,cid,sid,jid,did,pid,qty,value,), and:,C is the key:,C CSJDPQV,Project purchases each part using single contract:,JP C,Dept purchases at most one part from a supplier:,SD P,JP C, C CSJDPQV imply JP CSJDPQV,SD P implies SDJ JP,SDJ JP, JP CSJDPQV imply SDJ CSJDPQV,Inference Problems with Functional Dependencies,A set F of functional dependencies is given; does X,Y also hold (this is the same as saying is XY in F,+,),?,2 approaches can be used to answer this question:,Using the 3 (5) inference rules for functional dependencies see if you can derive X,Y,Compute the attribute closure of X, denoted by X,+,; if Y,X,+,; then XY holds; otherwise it doesnt hold (efficient!),Reasoning About FDs (Contd.),Computing the closure of a set of FDs can be expensive. (Size of closure is exponential in # attrs!),Typically, we just want to check if a given FD,X Y,is in the closure of a set of FDs,F,. An efficient check:,Compute,attribute closure,of X (denoted ) wrt,F:,Set of all attributes A such that X A is in,There is a linear time algorithm to compute this.,Check if Y is in,Does F = A B, B C, C D E imply A E?,i.e,is A E in the closure ? Equivalently, is E in ?,Using Axioms to Check if FD holds,Does F = A,B, B,C, C D,E imply A,E?,A,C,AD,DC,AD,E,Transitivity,Transitivity,Augmentation,AD,C,Decomposition,Remark,: many other FDs can be infered; however, we do not succeed in reaching A,E!,An Algorithm to Compute Attribute Closure X,+,with respect to F,Let X be a subset of the attributes of a relation R and F be the set of functional dependencies that hold for R.,Create a,hypergraph,in which the nodes are the attributes of the relation in question.,Create hyperlinks for all functional dependencies in F.,Mark all attributes belonging to X,Recursively continue marking unmarked attributes of the,hypergraph,that can be reached by a hyperlink with all ingoing edges being marked.,Result,: X,+,is the set of attributes that have been marked by this process.,Hypergraph for F,Does F = A,B, B,C, C D,E imply A,E?,A,B,C,D,E,Idea: Computer A,+;,if it contains E; A,E holds,The Evils of Redundancy,Redundancy,is at the root of several problems associated with relational schemas:,redundant storage, insert/delete/update anomalies,Integrity constraints, in particular,functional dependencies, can be used to identify schemas with such problems and to suggest refinements.,Main refinement technique:,decomposition,(replacing ABCD with, say, AB and BCD, or ACD and ABD).,Decomposition should be used judiciously:,Is there reason to decompose a relation?,What problems (if any) does the decomposition cause?,Example: A Bad Relational Design,Person,Works-,for,Company,(0,*),(0,*),ssn,name,salary,C#,loc,Table: X (ssn, name, salary, C#, loc),Insertion Anomaly,: Can we insert a person if they are not,working for a company,Deletion Anomaly,: If we delete the last employment of a company,we lose the information where the company is located,Update Anomaly,: If we change the city where a company is located,we have to update multiple,tuples,!,Example: Constraints on Entity Set,Consider relation obtained from Hourly_Emps:,Hourly_Emps (,ssn, name, lot, rating, hrly_wages,hrs_worked,),Notation,:,We will denote this relation schema by listing the attributes:,SNLRWH,This is really the,set,of attributes S,N,L,R,W,H.,Sometimes, we will refer to all attributes of a relation by using the relation name. (e.g., Hourly_Emps for SNLRWH),Some FDs on Hourly_Emps:,ssn,is the key:,S SNLRWH,rating,determines,hrly_wages,:,R W,Example (Contd.),Problems due to R W :,Update anomaly,:,Can we change W in just the 1st tuple of SNLRWH?,Insertion anomaly,:,What if we want to insert an employee and dont know the hourly wage for his rating?,Deletion anomaly,:,If we delete all employees with rating 5, we lose the information about the wage for rating 5!,Hourly_Emps2,Wages,Boyce-Codd Normal Form (BCNF),Reln R with FDs,F,is in,BCNF,if, for all X A in,A is a subset of X (called a,trivial,FD), or,X contains the attributes of a candidate key for R.,In other words, R is in BCNF if the only non-trivial FDs that hold over R are key constraints.,No dependency in R that can be predicted using FDs alone.,If we are shown two tuples that agree upon the X value, we cannot infer the A value in one tuple from the A value in the other.,If example relation is in BCNF, the 2 tuples must be identical (since X is a key).,What do we do a relation R is not in BCNF?,We decompose the relation R into smaller relations that are (hopefully) in BCNF,Example: R(A,B,C) with A,B. We decompose R into R1(A,B) with AB and R2(A,C) with no functional dependencies both of which are in BCNF,Question,: Should we also decompose R into R1 and R2, if R is not in BCNF and R1 and R2 are both in BCNF?,Decompositions: the Good and Bad News,Decompositions of “bad” functional dependencies reduce redundancy.,There are three potential problems to consider:,Some queries become more expensive.,Given instances of the decomposed relations, we may not be able to reconstruct the corresponding instance of the original relation (lossless join problem)!,Checking some dependencies may require joining the instances of the decomposed relations (problem of lost dependencies).,Tradeoff,:,Must consider these issues vs. redundancy.,Lossless Join Decompositions,Decomposition of R into X and Y is,lossless-join,w.r.t. a set of FDs F if, for every instance,r,that satisfies F:,(,r,) (,r,) =,r,It is always true that,r,(,r,) (,r,),In general, the other direction does not hold! If it does, the decomposition is lossless-join.,Definition extended to decomposition into 3 or more relations in a straightforward way.,It is essential that all decompositions used to deal with redundancy be lossless!,(Avoids Problem (2).),More on Lossless Join,The decomposition of R into X and Y is,lossless-join wrt F if,the closure of F contains:,X Y X, or,X Y Y,In particular, the decomposition of R into UV and R,-,V is lossless-join if U V holds over R.,Dependency Preserving Decomposition,Dependency preserving decomposition,(Intuitive): If R with attribute set Z is decomposed into X and Y, and we enforce the FDs that hold on X and on Y, then all FDs that were given to hold on Z must also hold.,(Avoids Problem (3).),Projection of set of FDs F,:,If Z is decomposed into X, . The projection of F onto X (denoted,F,X,) is the set of FDs,U V in F,+,(,closure of F,),such that,U, V subset of X,.,How to compute the F,X,?,(see Ullman book),Compute the attribute closure for every subset U of X;,If B in X, B in U,+, B not in U: add U B to F,X,.,Dependency Preserving Decompositions (Contd.),Decomposition of R into X and Y is,dependency,preserving,if,(F,X,union F,Y,),+,= F,+,i.e., if we consider only dependencies in the closure F,+,that can be checked in X without considering Y, and in Y without considering X, these imply all dependencies in F,+,.,Important to consider,F,+,not F, in this definition:,ABC, A B, B C, C A, decomposed into AB and BC.,Is this dependency preserving? Is C A preserved?,Dependency preserving does not imply lossless join:,ABC, A B, decomposed into AB and BC.,And vice-versa! (Example?),What is a “good” relational schema?,BCNF (or 4,th, 5,th, normal form),No lost functional dependencies,No unnecessary decompositions (minimum number of relations that satisfy the first and second condition).,Remark,: In same cases, conditions 1 and 2 cannot be jointly achieved.,Decomposition with respect to a functional dependency X, Y,Decompositions with respect to X,Y,:,Let R a relation with attributes ATT; furthermore, (X, Y)ATT, Z=ATT,-,(,X, Y) and X,Y holds and is non-trivial,In this case, R can be decomposed into R1 with attributes,X, Y and R2 with attributes,X, Z and,R1 R2=R (that is R can be reconstructed without loss of information).,Remark,: In the normalization process only decompositions with respect to a given functional dependency are used; from the above statement we know that all these decompositions are lossless.,Finding a “Good” Schema in BCNF,A relation R with ATT (R) =X and functional dependencies F is given,BCNF Decomposition Problem,: Find the smallest n and X,1,X,n,such that:,X,i,X for i=1,.,n,X,1,X,n,=X,R,i,with,ATT(R,i,)=X,i,and functional dependencies,F,i,is in BCNF for i=1,n,(F,1,F,n,),+,=F,+,(no lost functional dependencies),(R,1,|X| R,2,) |,X|R,n,)=R (|X|:= natural join),Remark,: Problem does not necessarily have a solution for certain relations R (e.g. R(A,B,C) with A,C and BC),Algorithm to find a “good” BCNF Relational Schema,Write down all (non-trivial) functional dependencies for the relation. Transform A,B1 and AB2 into AB1,B2,Identify the candidate keys of the relation,Classify functional dependencies into,Good,: have complete candidate key on their left-hand side,Bad,: not good,Compute all possible relational schemas using decompositions involving bad functional dependencies,Select the relational schema that is in BCNF and does not have any lost functional dependencies. If no such schema exists select a schema that comes closest to the ideal.,BCNF and Dependency Preservation,In general,there may not be a dependency preserving decomposition into BCNF,.,e.g., R(C,S,Z) CS Z, Z C,Cant decompose while preserving 1st FD; not in BCNF.,Summary of Schema Refinement,If a relation is in BCNF, it is free of redundancies that can be detected using FDs. Thus, trying to ensure that all relations are in BCNF is a good heuristic.,If a relation is not in BCNF, we can try to decompose it into a collection of BCNF relations.,Must consider whether all FDs are preserved.,Decompositions that do not guarantee the lossless-join property have to be avoided.,Decompositions should be carried out and/or re-examined while keeping,performance requirements,in mind.,Decompositions that do not reduce redundancy should be avoided.,