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Find the canonical cover of FD {A->BC, B->AC, C->AB} in DBMS
Canonical cover is called minimal cover which is called the minimum set of FDs. A set of FD FC is called canonical cover of F if each FD in FC is a Simple FD, Left reduced FD and Non-redundant FD.
Simple FD − X->Y is a simple FD if Y is a single attribute.
Left reduced FD: X->Y is a left reduced FD if there are no extraneous attributes in X.{extraneous attributes: let XA->Y then A is a extraneous attribute if X_>Y}
Non-redundant FD − X->Y is a Non-redundant FD if it cannot be derived from F- {X->y}.
Problem
Find the canonical cover of FD {A->BC, B->AC, C->AB}.
Solution
Relational schema R(A,B,C) F: {A->BC, B->AC, C->AB}
Step 1 − Create a singleton right hand side
dependency A->BC will break into A->B, A->C.
F: { A->B
A->C
B->A
B->C
C->A
C->B}Step 2 − Remove extraneous attributes if any exists.
F:{ A->B
A->C
B->A
B->C
C->A
C->B} NO extraneous attributes existsStep 3 − Remove the redundant FD
F: { A->B
A->C
B->A
B->C
C->A
C->B }Remove B->A dependency and we can get A from B through B->C and C->A.
F= {A->B
A->C
B->C
C->A
C->B}By removing C->B dependency we get B from C through C->A , A->B.
F={A->B
B->C
C->A
A->C}By removing A->C dependency we can determine C from A through A->B, B->C
Step 4 − The final canonical cover is as follows −
FC ={ A->B, B->C, C->A }
[A]+ =BC
[B]+=AC
[C]+=AB.
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