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Closure of an Attribute

Closure of an Attribute: Closure of an Attribute can be defined as a set of attributes that can be functionally determined from it.

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+, the closure of X under functional dependency set F

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

Input: A set F of FDs on a relation schema R, and a set of attributes X, which is a subset of R.

QUESTIONS ON CLOSURE SET OF ATTRIBUTE:

1) Given relational schema R( P Q R S T U V) having following attribute P Q R S T U and V, also there is a set of functional dependency denoted by FD = { P->Q, QR->ST, PTV->V }.

Determine Closure of (QR)+ and (PR)+

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.

Therefore QR+ = QRST (Answer)

Note: In FD PTV→V, T is in QRST but that cannot be entertained, as complete PTV should be a subset of QRST

b) 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 two FD that any left part of FD contains any P, R, Q, Since, in FD QR → ST has its complete left part QR in PQR

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.

Therefore PR+ = PRQST ( Answer)

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) Therefore T+ = TPQRS ( Answer).


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