【正文】 D, D224。E, CD224。B. We also found out that the keys are AB, AC and AD. FDs where the left side is not a superkey or the attributes on the right are not part of some key are 3NF violations. The 3NF violations are C224。B, ABCD224。B, ADE224。B, ACD224。C, ABD224。D, DE224。E, CD224。C, AD224。B, AC224。E, AB224。B, C224。C.Using algorithm , we can depose into relations using the minimal basis AB224。D, ABD224。D, BC224。C. We also found out that the only key is ABE. FDs where the left side is not a superkey or the attributes on the right are not part of some key are 3NF violations. The 3NF violations are AB224。C, BCE224。D, ABD224。D, BC224。A.We also found out that the keys are A,B,C,D. Since all the attributes on the right sides of the FDs are prime, there are no 3NF violations.Since there are no 3NF violations, it is not necessary to depose the relation.Exercise In the solution to Exercise we found that there are 16 nontrivial dependencies. They are AB224。B, ABC224。D, BD224。D, AD224。C, AB224。A, C224。A, A224。A.We also found out that the keys are AB, AD, BC, and CD. Since all the attributes on the right sides of the FDs are prime, there are no 3NF violations.Since there are no 3NF violations, it is not necessary to depose the relation.Exercise In the solution to Exercise we found that there are 28 nontrivial dependencies. They are A224。B, ABC224。B, AB224。D. The resulting deposed relations would be BC and BD. However, none of these two sets of attributes is a superkey. Thus we add relation AB to the result. The final set of deposed relations is BC, BD and AB. Exercise In the solution to Exercise we found that there are 12 nontrivial dependencies. They are AB224。D, BC224。C, ABC224。D, AB224。D, ABD224。D, BD224。D, AB224。D are not in the union and cannot be derived. Thus the two sets of FDs are not equivalent.Exercise In the solution to Exercise we found that there are 14 nontrivial dependencies. They are: C224。E are not in the union and cannot be derived. Thus the two sets of FDs are not equivalent.For the FDs of , the dependencies are not preserved. The union of the new sets of FDs is AC224。D. This is precisely the same as the original set of FDs and thus the two sets of FDs are equivalent.For the FDs of , the dependencies are not preserved. The union of the new sets of FDs is B224。DABCDEabcdea1bcdeab1cdeSince there is an unsubscripted row, the deposition for R is lossless for this set of FDs.Exercise When we depose a relation into BCNF, we will project the FDs onto the deposed relations to get new sets of FDs. These dependencies are preserved if the union of these new sets is equivalent to the original set of FDs. For the FDs of , the dependencies are not preserved. The union of the new sets of FDs is CE224。E and B224。A.ABCDEabcd1e1abcde1ab1cd1eSince there is not an unsubscripted row, the deposition for R is not lossless for this set of FDs.We can use the final tableau as an instance of R as an example for why the join is not lossless. The projected relations are:ABCabcab1cBCDbcd1bcdb1cd1ACEace1aceThe joined relation is:ABCDEabcd1e1abcde1ab1cd1e1abcd1eabcdeab1cd1eThe joined relation has three more tuples than the original tableau.Exercise This is the initial tableau:ABCDEabcd1e1a1bcde1ab1cd1eThis is the final tableau after applying FDs AC224。BC have A on the left side and part of the process of deposition involves finding {A}+ to form one deposed relation and A plus the rest of the attributes not in {A}+ as the second relation. Both cases yield the same deposed relations. Exercise Yes, we will still get the same result. Both A224。B. Using the above FDs, we get BDE and ABC as deposed relations. Using Algorithm to discover the projection of FDs on relation BDE, we discover that BDE is in BCNF since D, BD, DE are the only keys and all the projected FDs contain D, BD, or DE in the left side. Going back to relation ABC, following Algorithm tells us that ABC is not in BCNF because since AB and AC are its only keys and the FD C224。B, BCD224。B, CD224。E, BC224。B, C224。E, ABCE224。C, BCD224。E, ACE224。E, ABE224。B, ABC224。B, CD224。E, BC224。D, AC224。C, AB224。D, C224。D follows for ABCD. Using violation B224。C, BCE224。C, BE224。C, B224。C, ABCE224。C, ABE224。C, BE224。C, B224。C, ACD224。C, CD224。C, BC224。D, AC224。B, D224。D, B224。C, C224。C, ACD224。C, BC224。D, CD224。D and BD224。D and ABD224。C, AB224。D. Using the above FDs, we get ACD and BC as deposed relations. BC is surely in BCNF, since any twoattribute relation is. Using Algorithm to discover the projection of FDs on relation ACD, we discover that ACD is not in BCNF since C is its only key. However, D224。D, D224。D, ABD224。D, BD224。D, AB224。E. Since E is not in the closure, we can add attribute E.Given a set of FDs, we can prove that a FD F follows by taking the closure of the left side of FD F. The steps to pute the closure in Algorithm can be mimicked by Armstrong’s axioms and thus we can prove F from the given set of FDs using Armstrong’s axioms. Exercise In the solution to Exercise we found that there are 14 nontrivial dependencies, including the three given ones and eleven derived dependencies. They are: C224。ABD. We can use the splitting method in step one to get the FD ABC224。E.For steps two through four of Algorithm , suppose we have the initial set of attributes of the closure as ABC. Suppose also that we have FDs C224。DE and DE224。E follow. Surely the functional dependencies DE224。C and C224。B.Exercise We need to pute the closures of all subsets of {ABC}, although there is no need to think about the empty set or the set of all three attributes. Here are the calculations for the remaining six sets:{A}+=A{B}+=B{C}+=C{AB}+=ABD{AC}+=ABCDE{BC}+=ABCDEWe ignore D and E, so a basis for the resulting functional dependencies for ABC is: AC224。A, C224。A, C224。A, A224。A, B224。B, B224。DABD224。A BD224。B AD224。B D224。A C224。AExercise If the only closed sets are 248。BAB