【正文】 B and D224。C, C224。E and BCD224。E, BD224。E, D224。C and ACDE224。E, ABCE224。D, CDE224。C, BCD224。D, ADE224。E, ACE224。D, ACD224。E, ABE224。E, ABD224。B, ABC224。B, CE224。B, CD224。E, BD224。E, BC224。B, AD224。D, AC224。E, AC224。C, AB224。B, D224。D, C224。D. The resulting deposed relations would be ABC, CDE and BD. However, none of these three sets of attributes is a superkey. Thus we add relation ABE to the result. The final set of deposed relations is ABC, CDE, BD and ABE. Exercise In the solution to Exercise we found that there are 41 nontrivial dependencies. They are: C224。C, DE224。D and BDE224。C, ADE224。D, ABC224。D, BE224。D, AB224。C, DE224。D, and ABDE224。D, BDE224。D, ADE224。C, ABE224。D, ABC224。D, BE224。D, AB224。C, DE224。B and BCD224。D, ABD224。A, CD224。A, BD224。A, BC224。B, AD224。B, AC224。C, AB224。B, D224。A, C224。D, B224。C, A224。D, D224。B, B224。B and BCD224。D, ABD224。A, CD224。D, AD224。A, AD224。C, BC224。C and B224。D and BD224。C, B224。D and ABD224。D, BD224。C, AB224。C, B224。C, and BCD224。A, ABC224。A, BD224。A, BC224。 C, AC224。A, AB224。A, C224。E and E224。E. However, the FDs A224。D and D224。D and A224。E and BC224。A. However, the FD B224。E and E224。D.ABCDEabcdea1bcdeab1cdeSince there is an unsubscripted row, the deposition for R is lossless for this set of FDs.Exercise This is the initial tableau:ABCDEabcd1e1a1bcde1ab1cd1eThis is the final tableau after applying FDs A224。D, D224。E and BC224。E and CE224。B and A224。B and A224。B follows for ABC. Using violation C224。B.One choice is to depose using the violation D224。E, BCE224。D, DE224。E, CE224。E, CD224。D, BC224。B, D224。D, C224。B. From the closures we can also deduce that the keys are AB, AC and AD. Thus, any dependency above that does not contain one of the above pairs on the left is a BCNF violation. These are: C224。D, ABDE224。B, ABCD224。E, BCE224。B, ADE224。B, ACE224。B, ACD224。C, ABE224。C, ABD224。D, ABC224。D, DE224。E, CE224。E, CD224。D, BC224。C, AD224。E, AD224。B, AC224。D, AB224。E, AB224。E, D224。B, C224。D to further depose, we get BD and ABC as deposed relations. BD is in BCNF because it is a twoattribute relation. Using Algorithm again, we discover that ABC is in BCNF since AB is the only key and AB224。C. Using the above FDs, we get ABCD and ABE as deposed relations. Using Algorithm to discover the projection of FDs on relation ABCD, we discover that ABCD is not in BCNF since AB is its only key and the FD B224。D and BDE224。C, ADE224。D, ABC224。D, BE224。D, AB224。C, DE224。D, and ABDE224。D, BDE224。D, ADE224。C, ABE224。D, ABC224。D, BE224。D, AB224。C, DE224。B and BCD224。D, ABD224。A, CD224。A, BD224。A, BC224。B, AD224。B, AC224。C, AB224。B, D224。A, C224。D, B224。C, A224。D, D224。B, B224。B and BCD224。D, ABD224。A, CD224。D, AD224。A, AD224。C, BC224。C.One choice is to depose using the violation B224。D, BC224。C. From the closures we can also deduce that the only key is AB. Thus, any dependency above that does not contain AB on the left is a BCNF violation. These are: B224。C, ABC224。D, BC224。D, AB224。A is a dependency that holds in ABCD and therefore holds in ACD. We must further depose ACD into AD and CD. Thus, the three relations of the deposition are BC, AD, and CD.Exercise By puting the closures of all 15 nonempty subsets of ABCD, we can find all the nontrivial FDs. They are B224。A.One choice is to depose using the violation C224。A, AC224。A, C224。C, and BCD224。A, ABC224。A, BD224。A, BC224。 C, AC224。A, AB224。A, C224。ABCDE. Using again the splitting method in step one we get the FD ABCD224。D. Since D is not in the closure, we can add attribute D. Taking the FD D224。D and augmenting both sides with attributes AB we get the FD ABC224。D and D224。E and derive the FD ABC224。D. Likewise, we can do the same for ABC224。D from the FDs ABC224。D and DE224。D and ABC224。A.Exercise For step one of Algorithm , suppose we have the FD ABC224。B, B224。B and BC224。C. Note that BCA is true, but follows logically from CA, and therefore may be omitted from our list.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}+=AD{B}+=B{C}+=C{AB}+=ABDE{AC}+=ABCDE{BC}+=BCWe ignore D and E, so a basis for the resulting functional dependencies for ABC is: AC224。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}+=ACE{AB}+=ABCDE{AC}+=ACE{BC}+=ABCDEWe ignore D and E, so a basis for the resulting functional dependencies for ABC is: C224。C, C224。A}{A224。C, B224。C}{A224。B, B224。C, C224。B, B224。C, C224。AExercise We can think of this problem as a situation where the attributes A,B,C represent cities and the functional dependencies represent one way paths between the cities. The minimal bases are the minimal number of pathways that are needed to connect the cities. We do not want to create another roadway if the two cities are already connected.The systematic way to do this would be to check all possible sets of the pathways. However, we can simplify the situation by noting that it takes more than two pathways to visit the two other cities and e back. Also, if we find a set of pathways that is minimal, adding additional pathways will not create another minimal set.The two sets of minimal bases that were given in example are:{A224。CACD224。BABC224。CCD224。DBD224。CBC224。DAD224。CAC224。A D224。B C224。AC224。, {A,B} and {A,B,C,D}, then the following FDs hold:A224。BBCD224。DABD224。A CD224。A BD224。A BC224。B AD224。B AC224。C AB224。B D224。DD224。A C224。C B224。DB224。B A224。B is a nontrivial dependency. Then {A1A2...An}+ contains B and thus A1A2...An is not closed.Exercise If the only closed sets are 248。A1A2…Aj where A1A2…Aj is some subset of A1A2…An. It must be then that C1C2…Cm or some subset of C1C2…Cm is in X. However, the attributes C1C2…Cm cannot be in Y because we assumed that attributes A1A2…An are only in X+ and are not in Y+. Thus, X is not a subset of Y.By proving the contrap