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Tutorial3 - sol

# Tutorial3 - sol - Tutorial 3 Problem 1 BCNF form Consider a...

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Tutorial 3 Problem 1: BCNF form Consider a relation with schema R(A,B,C,D,E) and FD’s AB C, DE C, and B D. Is this relation in BCNF? Why or why not? If it's not, decompose the relation into collections of relations that are in BCNF. Make sure to indicate which dependency you apply to each decomposition . Solution Definition: A relation R is in BCNF if and only if every determinant (the left hand side of the FD) is a candidate key. R is not in BCNF. In fact, every FD is a violation. We decompose in steps: Step 1: AB C is a violate FD, (AB)+ is (ABCD) (recall that (AB)+ means the closure of AB). Therefore, we decompose R to R1(ABE) and R2(ABCD). We now examine R1 and R2. Step 2: R1 has no FDs, so it's in BCNF. Step 3: R2 has the FDs AB CD and B D. AB is a key, but B is not. Therefore we decompose R2 to R3(ABC) and R4(BD). Step 4: R3's FD is AB C and AB is a key, so it's in BCNF. Step 5: R4's FD is B

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Tutorial3 - sol - Tutorial 3 Problem 1 BCNF form Consider a...

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