CS 440 - Assignment #1 - Solutions
CS 440 ­ Assignment #1 Solutions (total = 150 pts., due on Feb. 1, Monday)
(10 pts.) Access Database Design View
Data view
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CS 440 - Assignment #1 - Solutions
(15 pts) Exercise 10.17 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
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CS 440 - Assignment #1 - Solutions
(15 pts) Exercise 10.18 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
0
0
0
0
0
1
1
0
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
0
0
1
0
1
1
X → Z, Y → Z are satisfied.
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CS 440 - Assignment #1 - Solutions
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
0
0
1
0
1
1
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
0
0
0
0
0
1
1
1
XY → Z, Z → W are satisfied.
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CS 440 - Assignment #1 - Solutions
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.
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CS 440 - Assignment #1 - Solutions
(15 pts) Exercise 10.21 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.
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CS 440 - Assignment #1 - Solutions
(15 pts) Exercise 10.28 on Page 374 in Elmasri and Navathe. a.
i. A → B: cannot hold. Tuples #1 and #2 cause the violation. t1[A] = t2[A] but t1[B] ≠ t2[B].
ii. B → C: may hold.
iii. C → B: cannot hold. Tuples #1 and #3 cause the violation. t1[C] = t3[C] but t1[B] ≠ t3[B].
iv. B → A: cannot hold. Tuples #1 and #5 cause the violation. t1[B] = t5[B] but t1[A] ≠ t5[A].
v. C → A: cannot hold. Tuples #1 and #3 cause the violation. t1[C] = t3[C] but t1[A] ≠ t3[A].
b. Yes, a potential candidate key is (A, B) or (A, C).
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CS 440 - Assignment #1 - Solutions
(15 pts) Exercise 10.32 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)
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CS 440 - Assignment #1 - Solutions
(15 pts) Exercise 10.35 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.
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