Solution Manual for First Course in Database Systems A 3E 3rd Edition

Solution Manual for First Course in Database Systems A 3E 3rd Edition

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FOR MORE OF THIS COURSE AND ANY OTHER COURSES, TEST BANKS, FINAL EXAMS, AND SOLUTION MANUALS CONTACT US AT [email protected] Exercise 3.1.1 Answers for this exercise may vary because of different interpretations. Some possible FDs: Social Security number name Area code state Street address, city, state zipcode Possible 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 phone Exercise 3.1.2 Answers for this exercise may vary because of different interpretations Some possible FDs: ID x-position, y-position, z-position ID x-velocity, y-velocity, z-velocity x-position, y-position, z-position ID Possible keys: {ID} {x-position, y-position, z-position} The reason why the positions would be a key is no two molecules can occupy the same point. Exercise 3.1.3a

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The superkeys are any subset that contains A 1 . Thus, there are 2 (n-1) such subsets, since each of the n-1 attributes A 2 through A n may independently be chosen in or out. Exercise 3.1.3b The superkeys are any subset that contains A 1 or A 2 . There are 2 (n-1) such subsets when considering A 1 and the n-1 attributes A 2 through A n . There are 2 (n-2) such subsets when considering A 2 and the n-2 attributes A 3 through A n . We do not count A 1 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 3.1.3c The superkeys are any subset that contains {A 1 ,A 2 } or {A 3 ,A 4 }. There are 2 (n-2) such subsets when considering {A 1 ,A 2 } and the n-2 attributes A 3 through A n . There are 2 (n-2) – 2 (n-4) such subsets when considering {A 3 ,A 4 } and attributes A 5 through A n along with the individual attributes A 1 and A 2 . We get the 2 (n-4) term because we have to discard the subsets that contain the key {A 1 ,A 2 } to avoid double counting. The total number of subsets is 2 (n-2) + 2 (n-2) – 2 (n-4) . Exercise 3.1.3d The superkeys are any subset that contains {A 1 ,A 2 } or {A 1 ,A 3 }. There are 2 (n-2) such subsets when considering {A 1 ,A 2 } and the n-2 attributes A 3 through A n . There are 2 (n-3) such subsets when considering {A 1 ,A 3 } and the n-3 attributes A 4 through A n We do not count A 2 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 3.2.1a 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 C A.
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