数据库系统基础教程第三章答案(共31页).doc
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1、精选优质文档-倾情为你奉上Exercise 3.1.1Answers for this exercise may vary because of different interpretations.Some possible FDs:Social Security number nameArea code stateStreet address, city, state zipcodePossible keys:Social Security number, street address, city, state, area code, phone numberNeed street addr
2、ess, 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 3.1.2Answers for this exercise may vary because of different interpretationsSome possible FDs:ID x-position, y
3、-position, z-positionID x-velocity, y-velocity, z-velocityx-position, y-position, z-position IDPossible keys:IDx-position, y-position, z-positionThe reason why the positions would be a key is no two molecules can occupy the same point.Exercise 3.1.3aThe superkeys are any subset that contains A1. Thu
4、s, there are 2(n-1) such subsets, since each of the n-1 attributes A2 through An may independently be chosen in or out.Exercise 3.1.3bThe superkeys are any subset that contains A1 or A2. There are 2(n-1) such subsets when considering A1 and the n-1 attributes A2 through An. There are 2(n-2) such sub
5、sets when considering A2 and the n-2 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(n-1) + 2(n-2).Exercise 3.1.3cThe superkeys are any subset that contains A1,A2 or A3,A4. There are 2(n-2)
6、 such subsets when considering A1,A2 and the n-2 attributes A3 through An. There are 2(n-2) 2(n-4) such subsets when considering A3,A4 and attributes A5 through An along with the individual attributes A1 and A2. We get the 2(n-4) term because we have to discard the subsets that contain the key A1,A2
7、 to avoid double counting. The total number of subsets is 2(n-2) + 2(n-2) 2(n-4).Exercise 3.1.3dThe superkeys are any subset that contains A1,A2 or A1,A3. There are 2(n-2) such subsets when considering A1,A2 and the n-2 attributes A3 through An. There are 2(n-3) such subsets when considering A1,A3 a
8、nd the n-3 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(n-2) + 2(n-3).Exercise 3.2.1aWe could try inference rules to deduce new dependencies until we are satisfied we have them all. A mor
9、e 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 CA.Now consider pairs of attributes: AB+ = ABCD, so we get new depe
10、ndency ABD. AC+ = ACD, and ACD is nontrivial. AD+ = AD, so nothing new. BC+ = ABCD, so we get BCA, and BCD. BD+ = ABCD, giving us BDA and BDC. CD+ = ACD, giving CDA.For the triples of attributes, ACD+ = ACD, but the closures of the other sets are each ABCD. Thus, we get new dependencies ABCD, ABDC,
11、and BCDA.Since ABCD+ = ABCD, we get no new dependencies.The collection of 11 new dependencies mentioned above are: CA, ABD, ACD, BCA, BCD, BDA, BDC, CDA, ABCD, ABDC, and BCDA. Exercise 3.2.1bFrom the analysis of closures above, we find that AB, BC, and BD are keys. All other sets either do not have
12、ABCD as the closure or contain one of these three sets.Exercise 3.2.1cThe 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 3.2.2ai) Fo
13、r the single attributes we have A+ = ABCD, B+ = BCD, C+ = C, and D+ = D. Thus, the new dependencies are AC and AD.Now consider pairs of attributes: AB+ = ABCD, AC+ = ABCD, AD+ = ABCD, BC+ = BCD, BD+ = BCD, CD+ = CD. Thus the new dependencies are ABC, ABD, ACB, ACD, ADB, ADC, BCD and BDC.For the trip
14、les of attributes, BCD+ = BCD, but the closures of the other sets are each ABCD. Thus, we get new dependencies ABCD, ABDC, and ACDB.Since ABCD+ = ABCD, we get no new dependencies.The collection of 13 new dependencies mentioned above are: AC, AD, ABC, ABD, ACB, ACD, ADB, ADC, BCD, BDC, ABCD, ABDC and
15、 ACDB.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 ABD, ADC, BCA and CDB.For the triples of attributes,
16、 all the closures of the sets are each ABCD. Thus, we get new dependencies ABCD, ABDC, ACDB and BCDA.Since ABCD+ = ABCD, we get no new dependencies.The collection of 8 new dependencies mentioned above are: ABD, ADC, BCA, CDB, ABCD, ABDC, ACDB and BCDA.iii) For the single attributes we have A+ = ABCD
17、, B+ = ABCD, C+ = ABCD, and D+ = ABCD. Thus, the new dependencies are AC, AD, BD, BA, CA, CB, DB and DC.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 co
18、llection of 24 new dependencies mentioned above are: AC, AD, BD, BA, CA, CB, DB, DC, ABC, ABD, ACB, ACD, ADB, ADC, BCA, BCD, BDA, BDC, CDA, CDB, ABCD, ABDC, ACDB and BCDA.Exercise 3.2.2bi) From the analysis of closures in 3.2.2a(i), we find that the only key is A. All other sets either do not have A
19、BCD as the closure or contain A.ii) From the analysis of closures 3.2.2a(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 3.2.2a(iii), we find that A, B, C and D are keys. All oth
20、er sets either do not have ABCD as the closure or contain one of these four sets.Exercise 3.2.2ci) The superkeys are all those sets that contain one of the keys in 3.2.2b(i). The superkeys are AB, AC, AD, ABC, ABD, ACD, BCD and ABCD.ii) The superkeys are all those sets that contain one of the keys i
21、n 3.2.2b(ii). The superkeys are ABC, ABD, ACD, BCD and ABCD.iii) The superkeys are all those sets that contain one of the keys in 3.2.2b(iii). The superkeys are AB, AC, AD, BC, BD, CD, ABC, ABD, ACD, BCD and ABCD.Exercise 3.2.3aSince A1A2AnC contains A1A2An, then the closure of A1A2AnC contains B. T
22、hus it follows that A1A2AnCB.Exercise 3.2.3bFrom 3.2.3a, we know that A1A2AnCB. Using the concept of trivial dependencies, we can show that A1A2AnCC. Thus A1A2AnCBC. Exercise 3.2.3cFrom A1A2AnE1E2Ej, we know that the closure contains B1B2Bm because of the FD A1A2An B1B2Bm. The B1B2Bm and the E1E2Ej
23、combine to form the C1C2Ck. Thus the closure of A1A2AnE1E2Ej contains D as well. Thus, A1A2AnE1E2EjD.Exercise 3.2.3dFrom A1A2AnC1C2Ck, we know that the closure contains B1B2Bm because of the FD A1A2An B1B2Bm. The C1C2Ck also tell us that the closure of A1A2AnC1C2Ck contains D1D2Dj. Thus, A1A2AnC1C2C
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