CS 440 Assignment #1 Solutions

Transcription

1 CS 440 Assignment #1 Solutions (total = 150 pts., due on Feb. 1, Monday) (10 pts.) Access Database Design View Data view 1

2 (15 pts) Exercise on Page 372 in Elmasri and Navathe. Student = {Sname, Snum, Ssn, Sc_addr, Sc_phone, Bdate, Sex, Class, Major_code, Minor_code, Prog} Ssn {Sname, Snum, Sc_addr, Sc_phone, Bdate, Sex, Class, Major_code, Prog} Snum {Sname, Ssn, Sc_addr, Sc_phone, Bdate, Sex, Class, Major_code, Prog} Department = {Dname, Dcode, Doffice, Dphone, Dcollege} Dname {Dcode, Doffice, Dphone, Dcollege} Dcode {Dname, Doffice, Dphone, Dcollege} Course = {Cname, Cdesc, Cnum, Credit, Level, Cdept} Cnum {Cname, Cdesc, Credit, Level, Cdept} Section = {Iname, Semester, Year, Sec_course, Sec_num} {Semester, Year, Sec_course, Sec_num} Iname Grade_record = {Ssn, Semester, Year, Sec_course, Sec_num, Grade} {Ssn, Semester, Year, Sec_course, Sec_num} Grade 2

3 (15 pts) Exercise on Page 373 in Elmasri and Navathe. a. Proof of {W Y, X Z} {WX Y} 1. W Y WX YX (using augmentation rule IR2) 2. YX Y (using reflexive rule IR1 and knowing that YX Y) 3. WX Y (using transitive rule IR3 on 1 and 2) b. Proof of {X Y} and Y Z {X Z} 1. X Y (given) 2. Y Z (using IR1 and knowing Y Z) 3. X Z (using IR3 on 1 and 2) c. Proof {X Y, X W, WY Z} {X Z} 1. X W (given) 2. XY WY (using IR2 on 1 by augmenting with Y) 3. X Y (given) 4. X XY (using IR2 on 1 by augmenting with X; notice that XX = X) 5. X WY (using IR3 on 2 and 4) 6. WY Z (given) 7. X Z (using IR3 on 4 and 5) d. Disproof of {XY Z, Y W} {XW Z} Consider the relation instance as below: X Y Z W XY Z and Y W are satisfied. But XW Z is not satisfied because with the same value (0, 0) of (X, W) in the first and the second record, the coresponding values of Z are different. e. Disproof of {X Z, Y Z} {X Y} Consider the relation instance as below: X Y Z X Z, Y Z are satisfied. 3

4 But X Y is not satisfied because with the same value 0 of X in the first and the second record, the coresponding values of Y are different. f. Proof of {X Y, XY Z} {X Z} 1. X Y (given) 2. X XY (using IR2 on 1 by augmenting with X; notice that XX = X) 3. XY Z (given) 4. X Z (using IR3 on 2 and 3) g. Proof of {X Y, Z W} {XZ YW} 1. X Y (given) 2. XZ YZ (using IR2 on 1 by augmenting with Z) 3. Z W (given) 4. YZ YW (using IR2 on 3 by augmenting with Y) 5. XZ YW (using IR3 on 2 and 4) h. Disproof of {XY Z, Z X} {Z Y} Consider the relation instance as below: X Y Z XY Z and Z X are satisfied. But Z Y is not satisfied because with the same value 0 of Z in the first and the second record, the coresponding values of Y are different. i. Proof of {X Y, Y Z} {X YZ} 1. Y Z (given) 2. Y YZ (using IR2 on 1 by augmenting with Y; notice that YY = Y) 3. X Y (given) 4. X YZ (using IR3 on 2 and 3) j. Disproof of {XY Z, Z W} {X W} Consider the relation instance as below: X Y Z W XY Z, Z W are satisfied. 4

5 But X W is not satisfied because with the same value 0 of X in the first and the second record, the coresponding values of W are different. 5

6 (15 pts) Exercise on Page 373 in Elmasri and Navathe. No, G is not minimal. Algorithm 10.2 (page 354) should be used to find a minimal set of functional dependencies G that is equivalent to G. G = {Ssn Ename, Ssn Bdate, Ssn Address, Ssn Dnumber, Dnumber Dname, Dnumber Dmgr_ssn} Definition of equivalent sets of functional dependencies (page 353) should be used to prove that G is equivalent to G. 6

7 (15 pts) Exercise on Page 374 in Elmasri and Navathe. a. i. A B: cannot hold. Tuples #1 and #2 cause the violation. t 1 [A] = t 2 [A] but t 1 [B] t 2 [B]. ii. B C: may hold. iii. C B: cannot hold. Tuples #1 and #3 cause the violation. t 1 [C] = t 3 [C] but t 1 [B] t 3 [B]. iv. B A: cannot hold. Tuples #1 and #5 cause the violation. t 1 [B] = t 5 [B] but t 1 [A] t 5 [A]. v. C A: cannot hold. Tuples #1 and #3 cause the violation. t 1 [C] = t 3 [C] but t 1 [A] t 3 [A]. b. Yes, a potential candidate key is (A, B) or (A, C). 7

8 (15 pts) Exercise on Page 375 in Elmasri and Navathe. * Normal Forms The relation is in 1NF because it does not have multivalued nor composite attributes. The relation is NOT in 2NF because the functional dependency Salesman# Commission% makes Commission% partially dependent on the primary key {Car#, Salesman#}, which violates the 2NF requirement. The relation is NOT in 3NF because of the transitive dependency of Discount_amt on the primary key via Date_sold. * Normalization The original CAR_SALE should be decomposed as below: 2NF: R1 (Car#, Date_sold, Salesman#, Discount_amt) R2 (Salesman#, Commission%) 3NF: R1 (Car#, Date_sold, Salesman#) R2 (Salesman#, Commission%) R3 (Date_sold, Discount_amt) 8

9 (15 pts) Exercise on Page 375 in Elmasri and Navathe. A E (given) AB E (augmentation rule) AB C (given) AB CE (union rule) CE D (given) AB D (transitive rule) That means AB D can be inferred from the the given set of dependencies. In other words, AB D is in the closure of the the given set of dependencies. 9