【正文】 Exercise Answers for this exercise may vary because of different interpretations.Some possible FDs: Social Security number 224。 name Area code 224。 state Street address, city, state 224。 zipcodePossible keys: {Social Security number, street address, city, state, area code, phone number}Need street address, city, state to uniquely determine location. A person could have multiple addresses. The same is true for phones. These days, a person could have a landline and a cellular phoneExercise Answers for this exercise may vary because of different interpretationsSome possible FDs: ID 224。 xposition, yposition, zposition ID 224。 xvelocity, yvelocity, zvelocity xposition, yposition, zposition 224。 IDPossible keys: {ID} {xposition, yposition, zposition}The reason why the positions would be a key is no two molecules can occupy the same point.Exercise The superkeys are any subset that contains A1. Thus, there are 2(n1) such subsets, since each of the n1 attributes A2 through An may independently be chosen in or out.Exercise The superkeys are any subset that contains A1 or A2. There are 2(n1) such subsets when considering A1 and the n1 attributes A2 through An. There are 2(n2) such subsets when considering A2 and the n2 attributes A3 through An. We do not count A1 in these subsets because they are already counted in the first group of subsets. The total number of subsets is 2(n1) + 2(n2).Exercise The superkeys are any subset that contains {A1,A2} or {A3,A4}. There are 2(n2) such subsets when considering {A1,A2} and the n2 attributes A3 through An. There are 2(n2) – 2(n4) such subsets when considering {A3,A4} and attributes A5 through An along with the individual attributes A1 and A2. We get the 2(n4) term because we have to discard the subsets that contain the key {A1,A2} to avoid double counting. The total number of subsets is 2(n2) + 2(n2) – 2(n4).Exercise The superkeys are any subset that contains {A1,A2} or {A1,A3}. There are 2(n2) such subsets when considering {A1,A2} and the n2 attributes A3 through An. There are 2(n3) such subsets when considering {A1,A3} and the n3 attributes A4 through An We do not count A2 in these subsets because they are already counted in the first group of subsets. The total number of subsets is 2(n2) + 2(n3).Exercise We could try inference rules to deduce new dependencies until we are satisfied we have them all. A more systematic way is to consider the closures of all 15 nonempty sets of attributes. For the single attributes we have {A}+ = A, {B}+ = B, {C}+ = ACD, and {D}+ = AD. Thus, the only new dependency we get with a single attribute on the left is C224。A.Now consider pairs of attributes: {AB}+ = ABCD, so we get new dependency AB224。D. {AC}+ = ACD, and AC224。D is nontrivial. {AD}+ = AD, so nothing new. {BC}+ = ABCD, so we get BC224。A, and BC224。D. {BD}+ = ABCD, giving us BD224。A and BD224。C. {CD}+ = ACD, giving CD224。A.For the triples of attributes, {ACD}+ = ACD, but the closures of the other sets are each ABCD. Thus, we get new dependencies ABC224。D, ABD224。C, and BCD224。A.Since {ABCD}+ = ABCD, we get no new dependencies.The collection of 11 new dependencies mentioned above are: C224。A, AB224。D, AC224。D, BC224。A, BC224。D, BD224。A, BD224。C, CD224。A, ABC224。D, ABD224。C, and BCD224。A. Exercise From the analysis of closures above, we find that AB, BC, and BD are keys. All other sets either do not have ABCD as the closure or contain one of these three sets.Exercise The superkeys are all those that contain one of those three keys. That is, a superkey that is not a key must contain B and more than one of A, C, and D. Thus, the (proper) superkeys are ABC, ABD, BCD, and ABCD.Exercise i) For the single attributes we have {A}+ = ABCD, {B}+ = BCD, {C}+ = C, and {D}+ = D. Thus, the new dependencies are A224。C and A224。D.Now consider pairs of attributes: {AB}+ = ABCD, {AC}+ = ABCD, {AD}+ = ABCD, {BC}+ = BCD, {BD}+ = BCD, {CD}+ = CD. Thus the new dependencies are AB224。C, AB224。D, AC224。B, AC224。D, AD224。B, AD224。C, BC224。D and BD224。C.For the triples of attributes, {BCD}+ = BCD, but the closures of the other sets are each ABCD. Thus, we get new dependencies ABC224。D, ABD224。C, and ACD224。B.Since {ABCD}+ = ABCD, we get no new dependencies.The collection of 13 new dependencies mentioned above are: A224。C, A224。D, AB224。C, AB224。D, AC224。B, AC224。D, AD224。B, AD224。C, BC224。D, BD224。C, ABC224。D, ABD224。C and ACD224。B.ii) For the single attributes we have {A}+ = A, {B}+ = B, {C}+ = C, and {D}+ = D. Thus, there are no new dependencies.Now consider pairs of attributes: {AB}+ = ABCD, {AC}+ = AC, {AD}+ = ABCD, {BC}+ = ABCD, {BD}+ = BD, {CD}+ = ABCD. Thus the new dependencies are AB224。D, AD224。C, BC224。A and CD224。B.For the triples of attributes, all the closures of the sets are each ABCD. Thus, we get new dependencies ABC224。D, ABD224。C, ACD224。B and BCD224。A.Since {ABCD}+ = ABCD, we get no new dependencies.The collection of 8 new dependencies mentioned above are: AB224。D, AD224。C, BC224。A, CD224。B, ABC224。D, ABD224。C, ACD224。B and BCD224。A.iii) For the single attributes we have {A}+ = ABCD, {B}+ = ABCD, {C}+ = ABCD, and {D}+ = ABCD. Thus, the new dependencies are A224。C, A224。D, B224。D, B224。A, C224。A, C224。B, D224。B and D224。C.Since all the single attributes’ closures are ABCD, any superset of the single attributes will also lead to a closure of ABCD. Knowing this, we can enumerate the rest of the new dependencies.The collection of 24 new dependencies mentioned above are: A224。C, A224。D, B224。D, B224。A, C224。A, C224。B, D224。B, D224。C, AB224。C, AB224。D, AC224。B, AC224。D, AD224。B, AD224。C, BC224。A, BC224。D, BD224。A, BD224。C, CD224。A, CD224。B, ABC224。D, ABD224。C, ACD224。B and BCD224。A.Exercise i) From the analysis of closures in (i), we find that the only key is A. All other sets either do not have ABCD as the closure or contain A.ii) From the analysis of closures (ii), we find that AB, AD, BC, and CD are keys. All other sets either do not have ABCD as the closure or contain one of these four sets.iii) From the analysis of closures (iii), we find that A, B, C and D