Question

Dependency Very Good dependency (key dependency) : XA where Table Very good If all dependencies in a table are '"very good", the table is in BCNF X is a candidate key Good Good dependency: X-> A where If all dependencies in a table are "very good" or "good", the table is in 3NF X is not a candidate key X is a part of a candidate key A is prime attribute Bad Bad dependency (Transitive Dependency): X A where If all dependencies in a table are "very good", "good", or bad, the table is in 2NF X is not a candidate key X is not a part of a candidate key A is non-prime attribute Very bad Very Bad dependency (Partial Dependency):X> A where Otherwise There is at least one very bad, the table is in INF X is a part of a candidate key A is non-prime attribute Problem Bonus Consider the relation R(P,Y,X,Z,E) with FD set (P>Z, Y->P, YX->Z, PX-YE) 1. Find all possible candidate keys of the relation 2. Find the Prime attributes and non-prime attributes 3. Classify each functional dependency 4. Determine the Normal Form of the relation 5. Decompose the relation into BCNF relation. 6. Determine whether each of the decompositions has the lossless join property with respect to FDs and conclude for the whole decomposition 7. Determine whether each of the decompositions has the dependency preserving property with respect to FDs and conclude for the whole decomposition

Answer 1

sien Relaton R (P,y,x,7,E) (t) find au tosible Candidale res closurs of P P, Z În the given FD set F , NO FD can dekumine. Aath' hubcX So auribhue x must includc in Candidak Key , (i.e. ei br Subset of Kk So , Px is a candida ke^ . Candi dale key, or 1f c deer mines AB then c ỉs ack. so , candidate key set = 〈A, yx } Find the prime & non-prime atributesha since, Candi dale t»y seh Non-Prime attsibuez,EJ

<ince, PX ,YX are candidale re veraood depondencies ropes subser of candidaler"J, PX and Z ĺç non-pr1 rre, bad depender since, P ís then XY→ P All FD in the table are not verj good, sa elation R is ot in BCNF. since, we got (p>z) in dependenc so, Tclaltion R can not de in 2NF. Since , relation R does not cord an ar번 multi valued aHr81 bule, So, Relation R is in 1 NE *Note: According to chaJ& GuideLines , I have osted solution oA first fouspas of ha question. Please in separale pot