Using this heuristic (see: click here), the necessary attributes are: A B (they are missing on the RHS) When we compute the closure of AB, we find that: AB+ = A B = A B C (A,B --> C) = A B C D E (A --> D) = A B C D E F (B --> F) = A B C D E F G H (F --> G H) = A B C D E F G H I J (D --> I J) = whole relation R

Decomposition into BCNF:

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)

Using this heuristic (see: click here), the necessary attributes are: A B D (they are missing on the RHS) When we compute the closure of ABD, we find that: ABD+ = ABD C E F G H I J = R

Therefore: key of R = ABD

Decomposition into BCNF:

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) <-- trivial relation, it is OK to omit the trivial relation...

   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).

    AB+ = A B C
    CD+ = C D E B
    DE+ = D E B

    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