## Closure of an Attribute
OR Closure of a set F of FDs is the set F+ of all FDs that can be inferred from F Closure of a set of attributes X concerning F is the set X+ of all attributes that are functionally determined by X ## Pseudocode to find Closure of an Attribute?Determine X X Closure : = will contain X itself; Repeat the process as: old X Closure : = X Closure; for each functional dependency P → Q in FD set do if X Closure is subset of P then X Closure := X Closure U Q ; Repeat until ( X Closure = old X Closure); ## Algorithm of Determining X+, the Closure of X under F
## QUESTIONS ON CLOSURE SET OF ATTRIBUTE:1) Given relational schema Determine Closure of a) QR+ = QR (as the closure of an attribute or set of attributes contain same). Now as per algorithm look into a set of FD that complete the left side of any FD contains either Q, R, or QR since in FD QR→ST has complete QR. Hence QR+ = QRST Again, trace the remaining two FD that any left part of FD contains any Q, R, S, T. Since no complete left side of the remaining two FD{P->Q, PTV->V} contain Q, R, S, T.
## Note: In FD PTV→V, T is in QRST but that cannot be entertained, as complete PTV should be a subset of QRSTb) PR + = PR (as the closure of an attribute or set of attributes contain same) Now as per algorithm look into a set of FD, and check that complete left side of any FD contains either P, R, or PR. Since in FD P→Q, P is a subset of PR, Hence PR+ = PRQ Again, trace the remaining Hence PR+ = PRQST Again trace the remaining one FD { PTV->V } that its complete left belongs to PRQST. Since complete PTV is not in PRQST, hence we ignore it.
2. Given relational schema R( P Q R S T) having following attributes P Q R S and T, also there is a set of functional dependency denoted by FD = { P->QR, RS->T, Q->S, T-> P }. Determine Closure of ( T ) T + = T (as the closure of an attribute or set of attributes contain same) Now as per algorithm look into a set of FD that complete the left side of any FD contains T since, in FD T → P, T is in T, Hence T+ = TP Again trace the remaining three FD that any left part of FD contain any TP, Since in FD P → QR has its complete left part P in TP, Hence T+ = TPQR Again trace the remaining two FD { RS->T, Q->S } that any of its Complete left belongs to TPQR, Since in FD Q → S has its complete left part Q in TPQR, Hence T+ = TPQRS Again trace the remaining one FD { RS->T } that its complete left belongs to TPQRS, Since in FD RS → T has its complete left part RS in TPQRS Hence T+ = TPQRS ( no changes, as T, is already in TPQRS) Next TopicQUESTIONS ON BOYCE CODD NORMAL FORM |