Functional Dependency & Decomposition Homework Solution

- Key of relation
`R`: AB

Because AB^{+}= ABCDEFGHIJ

But A^{+}= ADEIJ and B^{+}= BFGHDecomposition:

ABCDEFGHIJ Violation: A -> DE ADEIJ (A

^{+}) key = A Violation: D -> IJ DIJ (D^{+}) key = D ADE key = A ABCFGH key = AB Violation: B -> F BFGH (B^{+}) Violation: F -> GH FGH (F^{+}) key = F BF key = B ABC**Final answer: (D I J) (A D E) (F G H) (B F) (A B C)** - Key of relation
`R`: ABDDecomposition:

ABCDEFGHIJ Violation: AB -> C ABCI (AB

^{+}) key = AB Violation: A -> I AI (A^{+}) key = A ABC key = AB ABDEFGHJ key = ABD Violation: BD -> EF BDEF (BD^{+}) key = BD ABDGHJ key = ABD Violation: AD -> GH ADGHJ (AD^{+}) key = AD Violation: H -> J HJ (H^{+}) key = H ADGH key = AD ABD key = ABD (trivial relation)**Final answer: (A I) (A B C) (B D E F) (H J) (A D G H) (A B D) - it is OK to omit the trivial relation...** - Given:
Tuple# A B C ---------------------------------------- #1 10 b1 c1 #2 10 b2 c2 #3 11 b4 c1 #4 12 b3 c4 #5 13 b1 c1 #6 14 b3 c4 1. A -> B CANNOT hold because 2 different B values for same value A=10: (10

**b1**c1) and (10**b2**c2) 2. B -> C may hold 3. C -> B CANNOT hold because 2 different B values for same value C=c1: (10**b1**c1) and (11**b4**c1) 4. B -> A CANNOT hold because 2 different A values for same value B=b1: (**10**b1 c1) and (**13**b1 c1) 5. C -> A CANNOT hold because 2 different A values for same value C=c1: (**10**b1 c1) and (**13**b1 c1) Potential keys: AB and AC BC is not a potential key because of the tuples: (10 b1 c1) and (13 b1 c1). - The closures are:
AB

^{+}= A B C CD^{+}= C D E B DE^{+}= D E B - AD must be part of any key
AD

^{+}= A D Try adding B: ABD^{+}= A B D C E**OK**Try adding C: ACD^{+}= A C D E B**OK**Try adding E: ADE^{+}= A D E B C**OK**All keys: ABD, ACD and ADE