Equivalence of Sets of Functional Dependencies Example –
- Let F & G are two functional dependency sets. These two sets F & G are equivalent if E+=F+. Equivalence means that every functional dependency in F can be inferred from G, and every functional dependency in G an be inferred from F. →
- F and G are equal only if
- F covers G- means that all functional dependency of G are logically numbers of functional dependency set F⇒F⊇G.
- G covers F-means that all functional dependency of F are logically members of functional dependency set G⇒G⊇F
| F covers G | True | True | False | False |
| G covers F | True | False | True | False |
| Result | F=G | F⊃G | G⊃F | No Comparison |
Question 1 : Consider the two sets F and G with their FDs as below : F : G: A → C A → CD AC → D E → AH E → AD E → H Check whether two sets are equivalent or not.
Solution :
Step 1 : Take Set F and Check G is covered from F or not.
(A)+ = {ACD}
(E)+ = {EADHC}
Hence, both A → CD and E → AH are covered.
⇒ G is derived from F. Hence G is covered by F.
⇒ F ⊇ G . ....(1)
Step 2 : Take Set G and Check F is covered from G or not.
(A)+ = {ACD}
(AC)+ = {ACD}
(E)+ = {EAHCD}
Hence F = {A → C, AC → D, E → AD, E → H} is covered.
⇒ F is derived from G. Hence F is covered from G.
⇒ G ⊇ F. ....(2)
From (1) and (2), F and G are equivalent.
Question 2 : Consider the two sets P and Q with their FDs as below : P : Q : A → B A → BC AB → C D → AE D → ACE Check whether two sets are equivalent or not.
Solution :
Step 1 : Take Set P and Check Q is covered from P or not.
(A)+ = {ABC}
(D)+ = {DACEB}
Hence, both A → BC and D → AE are covered.
⇒ Q is derived from P. Hence Q is covered by P.
⇒ P ⊇ Q . ....(1)
Step 2 : Take Set Q and Check P is covered from Q or not.
(A)+ = {ABC}
(AB)+ = {ABC}
(D)+ = {DAEBC}
Hence P = {A → B, AB → C, D → ACE} is covered.
⇒ P is derived from Q. Hence P is covered by Q.
⇒ Q ⊇ P. ....(2)
From (1) and (2), P and Q are equivalent.
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